Ecological Effects of Predator Harvesting and Environmental Noises on Oceanic Coral Reefs

Abstract

Coral reefs provide refuge for prey and are important for the preservation of an oceanic ecosystem. However, they have been experiencing severe destruction by environmental changes and human activities. In this paper, we propose and analyze a tri-trophic food chain model consisting of coral, Crown-of-thorns starfish (CoTS), and triton in deterministic and stochastic environments. We investigate the effects of harvesting in the deterministic system and environmental noises in the stochastic system, respectively. The existence of possible steady states along with their stability is rigorously discussed. From the economic perspective, we examine the existence of the bionomic equilibrium and establish the optimal harvesting policy. Subsequently, the deterministic system is extended to a stochastic system through nonlinear perturbation. The stochastic system admits a unique positive global solution initiating from the interior of the positive quadrant. The long-time behaviors of the stochastic system are explored. Numerical simulations are provided to validate and complement our theoretical results. We show that over-harvesting of triton is not beneficial to coral reefs and modest harvesting of CoTS may promote sustainable growth in coral reefs. In addition, the presence of strong noises can lead to population extinction.

Appendices

Appendix A: Proofs for Deterministic System

1.1 Proof of Lemma 1

Proof

According to the first equation of (1), we have

$$\begin{aligned} x(t)=x(0)\textrm{exp}\left( \int _{0}^{t}\left[ r_{1}\left( 1-\frac{s}{K_{1}}\right) -\frac{m_{1}y}{a_{1}+s}\right] \textrm{d}s\right) . \end{aligned}$$

Similarly, we obtain

$$\begin{aligned} y(t)=y(0)\textrm{exp}\left( \int _{0}^{t}\left[ r_{2}\left( 1-\frac{s}{K_{2}}\right) +\frac{e_{1}m_{1}x}{a_{1}+x}-\frac{m_{2}z}{a_{2}+s}-h_{1}\right] \textrm{d}s\right) \end{aligned}$$

and

$$\begin{aligned} z(t)=z(0)\textrm{exp}\left( \int _{0}^{t}\left( \frac{e_{2}m_{2}y}{a_{2}+y}-d-h_{2}\right) \textrm{d}s\right) . \end{aligned}$$

It is clear that $x(t)>0$, $y(t)>0$, and $z(t)>0$ whenever $x(0)>0$, $y(0)>0$, and $z(0)>0$. Then any solution from the first quadrant of xyz-plane gives a positive solution. This completes the proof. $\square$

1.2 Proof of Theorem 1

Proof

Define $W=e_{1}e_{2}x+e_{2}y+z$. Taking time derivative of W leads to

$$\begin{aligned} \frac{\textrm{d}W}{\textrm{d}t}+(d+h_{2})W= & {} e_{1}e_{2}r_{1}x\left( 1-\frac{x}{K_{1}}\right) +e_{2}r_{2}y\left( 1-\frac{y}{K_{2}}\right) -e_{2}h_{1}y-dz-h_{2}z\\{} & {} \quad \displaystyle +(d+h_{2})(e_{1}e_{2}x+e_{2}y+z)\\\le & {} -\frac{e_{1}e_{2}r_{1}}{K_{1}}\left( x-\frac{K_{1}(r_{1}+d+h_{2})}{2r_{1}}\right) ^{2}+\frac{e_{1}e_{2}K_{1}(r_{1}+d+h_{2})^{2}}{4r_{1}}\\{} & {} \quad \displaystyle -\frac{e_{2}r_{2}}{K_{2}}\left( y-\frac{K_{2}(r_{2}+d+h_{2})}{2r_{2}}\right) ^{2}+\frac{e_{2}K_{2}(r_{2}+d+h_{2})^{2}}{4r_{2}}\\\le & {} \frac{e_{1}e_{2}K_{1}(r_{1}+d+h_{2})^{2}}{4r_{1}}+\frac{e_{2}K_{2}(r_{2}+d+h_{2})^{2}}{4r_{2}}\\:= & {} \mu . \end{aligned}$$

Due to the positivity of parameters and the nonnegativity of solutions, we obtain

$$\begin{aligned} 0\le W(x,y,z)\le \frac{\mu }{d+h_{2}}(1-\textrm{e}^{-(d+h_{2})t})+W(x(0),y(0),z(0))\textrm{e}^{-(d+h_{2})t}. \end{aligned}$$

Taking the limit of the above inequality as t tends to $\infty$, we obtain

$$\begin{aligned} \limsup _{t\rightarrow \infty }W(x,y,z)\le \frac{\mu }{d+h_{2}}. \end{aligned}$$

Therefore, all solutions of (1) with initial values $(x(0),y(0),z(0))\in \mathbb {R}^{3}_{+}Z$ are eventually confined to the region

$$\begin{aligned} \Omega =\left\{ (x,y,z)\in \mathbb {R}^{3}_{+}:~0\le e_{1}e_{2}x+e_{2}y+z\le \frac{\mu }{d+h_{2}}+1\right\} . \end{aligned}$$

$\square$

1.3 Proof of Lemma 2

Proof

When $n=1$, it is obvious that the claim is valid. Assume that the claim holds for $n=2k-1$. Then, when $n=2k+1$, one has

$$\begin{aligned} \alpha (x)=(-1)^{2k+1}x^{2k+1}+p_{2k}x^{2k}+\cdots +p_{1}x+p_{0}. \end{aligned}$$

If $\alpha (x)=0$ has one positive root, then the claim is valid. If all the roots are negative, then there is a contradiction with $p_0>0$. Thus, $\alpha (x)=0$ has at least a pair of conjugate complex roots denoted by $x_0=a+bi, \bar{x}_0=a-bi$ with $a\not =0$. Then,

$$\begin{aligned} \begin{aligned} \alpha (x)&=(x-a-bi)(x-a+bi)\tilde{\alpha }(x)=(x^{2}-2ax+a^{2}+b^{2})\tilde{\alpha }(x), \end{aligned} \end{aligned}$$

where $\tilde{\alpha }(x)=(-1)^{2k-1}x^{2k-1}+q_{2k-2}x^{2k-2}+\cdots +q_{1}x+q_{0}$ and $q_{0}(a^{2}+b^{2})=p_{0}>0$. By the method of mathematical induction, $\alpha (x)$ has at least one positive root if $p_{0}>0$. The proof is complete. $\square$

1.4 Proof of Theorem 2

Proof

The local stability of equilibria is determined by computing the eigenvalues of the Jacobian matrix about each equilibrium. The Jacobian matrix of (1) at $E_{0}$ is

$$\begin{aligned} J\mid _{E_{0}}=\left( \begin{array}{cccccc} r_{1}&{}0&{}0\\ 0&{}r_{2}-h_{1}&{}0\\ 0&{}0&{}-d-h_{2}\\ \end{array} \right) . \end{aligned}$$

The three eigenvalues of this matrix are

$$\begin{aligned} \lambda _{1}=r_{1},~~\lambda _{2}=r_{2}-h_{1},~~\lambda _{3}=-d-h_{2}. \end{aligned}$$

Obviously, $E_{0}=(0,0,0)$ is unstable. The proof is complete. $\square$

1.5 Proof of Theorem 3

Proof

The Jacobian matrix of (1) at $E_{x}$ reads

$$\begin{aligned} J\mid _{E_{x}}=\left( \begin{array}{cccccc} -r_{1}&{}-\frac{m_{1}K_{1}}{a_{1}+K_{1}}&{}0\\ 0&{}r_{2}+\frac{e_{1}m_{1}K_{1}}{a_{1}+K_{1}}-h_{1}&{}0\\ 0&{}0&{}-d-h_{2}\\ \end{array} \right) . \end{aligned}$$

The three eigenvalues of this matrix are

$$\begin{aligned} \lambda _{1}=-r_{1},~~\lambda _{2}=r_{2}+\frac{e_{1}m_{1}K_{1}}{a_1+K_{1}}-h_{1},~~\lambda _{3}=-d-h_{2}. \end{aligned}$$

The equilibrium $E_{x}=(K_{1},0,0)$ is locally asymptotically stable if $h_{1}>r_{2}+{e_{1}m_{1}K_{1}}/({a_{1}+K_{1}})$, otherwise it is a saddle point with stable manifold in $x-z$ plane and unstable manifold in y direction. The proof is complete. $\square$

1.6 Proof of Theorem 4

Proof

The Jacobian matrix of (1) at $E_{y}$ takes the form

$$\begin{aligned} J\mid _{E_{y}}=\left( \begin{array}{cccccc} r_{1}-\frac{m_{1}y_{2}}{a_{1}}&{}0&{}0\\ \frac{e_{1}m_{1}y_{2}}{a_{1}}&{}-r_{2}+h_{1}&{}-\frac{m_{2}y_{2}}{a_{2}+y_{2}}\\ 0&{}0&{}\frac{e_{2}m_{2}y_{2}}{a_{2}+y_{2}}-d-h_{2}\\ \end{array} \right) . \end{aligned}$$

Its eigenvalues are

$$\begin{aligned} \lambda _{1}=r_{1}-\frac{m_{1}K_{2}(r_{2}-h_{1})}{a_{1}r_{2}},~~\lambda _{2}=-(r_{2}-h_{1}), \\ \lambda _{3}=\frac{e_{2}m_{2}K_{2}(r_{2}-h_{1}) -[a_{2}r_{2}+K_{2}(r_{2}-h_{1})](d+h_{2})}{a_{2}r_{2}+K_{2}(r_{2}-h_{1})}. \end{aligned}$$

Then it is not difficult to show that all the eigenvalues of $J\mid _{E_{y}}$ have negative real parts when (6) is satisfied, then $E_{y}$ is locally asymptotically stable. The proof is complete. $\square$

1.7 Proof of Theorem 5

Proof

By (3), one has

$$\begin{aligned} y_{3}=\frac{(a_{1}+x_{3})(r_{1}K_{1}-r_{1}x_{3})}{m_{1}K_{1}}. \end{aligned}$$

Substituting the above equation into the second equation of (3) shows that $x_{3}$ is a positive solution of the following equation:

$$\begin{aligned} -x^{3}+(K_{1}-2a_{1})x^{2}+\hat{\nu }x +a_{1}^{2}K_{1}-\frac{a_{1}m_{1}K_{1}K_{2}(r_{2}-h_{1})}{r_{1}r_{2}}=0, \end{aligned}$$

where $\hat{\nu }=[{m_{1}K_{1}K_{2}(h_{1}-r_{2}-e_{1}m_{1})+a_{1}r_{1}r_{2}(2K_{1}-a_{1})}]/(r_{1}r_{2})$. From Lemma 2, it follows that $x_{3}$ is positive if (7) is satisfied.

The Jacobian matrix of (1) at $E_{xy}$ writes

$$\begin{aligned} J\mid _{E_{xy}}=\left( \begin{array}{cccccc} \frac{m_{1}x_{3}y_{3}}{(a_{1}+x_{3})^{2}}-\frac{r_{1}}{K_{1}}x_{3}&{}-\frac{m_{1}x_{3}}{a_{1}+x_{3}}&{}0\\ \frac{e_{1}m_{1}a_{1}y_{3}}{(a_{1}+x_{3})^{2}}&{}-\frac{r_{2}}{K_{2}}y_{3}&{}-\frac{m_{2}y_{3}}{a_{2}+y_{3}}\\ 0&{}0&{}\frac{e_{2}m_{2}y_{3}}{a_{2}+y_{3}}-d-h_{2}\\ \end{array} \right) . \end{aligned}$$

The characteristic equation of $J\mid _{E_{xy}}$ of (1) is

$$\begin{aligned} \left[ \lambda ^{2}-(J_{11}+J_{22})\lambda +J_{11}J_{22}-J_{12}J_{21}\right] \left[ \lambda -\left( \frac{e_{2}m_{2}y_{3}}{a_{2}+y_{3}}-d-h_{2}\right) \right] =0. \end{aligned}$$

Then $\lambda _{1}<0,~\lambda _{2}<0$, and $\lambda _{3}<0$ under the conditions of Theorem 5. Thus, $E_{xy}=(x_{3},y_{3},0)$ is locally asymptotically stable. The proof is complete. $\square$

1.8 Proof of Theorem 6

Proof

By (4), one has

$$\begin{aligned} y_{4}=\frac{a_{2}(d+h_{2})}{e_{2}m_{2}-d-h_{2}},~~z_{4}=\frac{(a_{2}+y_{4})[K_{2}(r_{2}-h_{1})-r_{2}y_{4}]}{m_{2}K_{2}}. \end{aligned}$$

It is trivial to show that $E_{yz}=\left( 0,y_{4},z_{4}\right)$ exists when (8) is satisfied.

The Jacobian matrix of (1) evaluated at $E_{yz}$ reads

$$\begin{aligned} J\mid _{E_{yz}}=\left( \begin{array}{cccccc} r_{1}-\frac{m_{1}y_{4}}{a_{1}}&{}0&{}0\\ \frac{e_{1}m_{1}y_{4}}{a_{1}}&{}\dfrac{m_{2}y_{4}z_{4}}{(a_{2}+y_{4})^{2}}-\frac{r_{2}}{K_{2}}y_{4}&{}-\frac{m_{2}y_{4}}{a_{2}+y_{4}}\\ 0&{}\frac{e_{2}m_{2}a_{2}z_{4}}{(a_{2}+y_{4})^{2}}&{}0\\ \end{array} \right) . \end{aligned}$$

The characteristic equation of $J\mid _{E_{yz}}$ is

$$\begin{aligned} \left[ \lambda ^{2}-\left( \frac{m_{2}y_{4}z_{4}}{(a_{2}+y_{4})^{2}}-\frac{r_{2}}{K_{2}}y_{4}\right) \lambda +\frac{e_{2}a_{2}m_{2}^{2}y_{4}z_{4}}{(a_{2}+y_{4})^{3}}\right] \left[ \lambda -\left( r_{1}-\frac{m_{1}y_{4}}{a_{1}}\right) \right] =0. \end{aligned}$$

Then, $\lambda _{1}<0$, $\lambda _{2}<0$, and $\lambda _{3}<0$ when (9) holds. Therefore, $E_{yz}=(0,y_{4},z_{4})$ is locally asymptotically stable. The proof is complete. $\square$

1.9 Proof of Theorem 7

Proof

It is not difficult to show that $y^*>0$ and $z^*>0$ when (11) is satisfied. The existence of $E_{xyz}(x^{*},y^{*},z^{*})$ follows from direct calculation. The Jacobian matrix of (1) evaluated at $E_{xyz}$ is

$$\begin{aligned} J\mid _{E_{xyz}}=\left( \begin{array}{cccccc} \frac{m_{1}x^{*}y^{*}}{(a_{1}+x^{*})^{2}}-\frac{r_{1}}{K_{1}}x^{*}&{}-\frac{m_{1}x^{*}}{a_{1}+x^{*}}&{}0\\ \frac{e_{1}m_{1}a_{1}y^{*}}{(a_{1}+x^{*})^{2}}&{}\frac{m_{2}y^{*}z^{*}}{(a_{2}+y^{*})^{2}}-\frac{r_{2}}{K_{2}}y^{*}&{}-\frac{m_{2}y^{*}}{a_{2}+y^{*}}\\ 0&{}\frac{e_{2}m_{2}a_{2}z^{*}}{(a_{2}+y^{*})^{2}}&{}0\\ \end{array} \right) . \end{aligned}$$

The characteristic equation of $J\mid _{E_{xyz}}$ of (1) is

$$\begin{aligned} \lambda ^{3}+A_{1}\lambda ^{2}+A_{2}\lambda +A_{3}=0, \end{aligned}$$

where

$$\begin{aligned} A_{1}=-(J_{11}+J_{22}),~~A_{2}=J_{11}J_{22}-J_{23}J_{32}-J_{12}J_{21},~~A_{3}=J_{11}J_{23}J_{32}. \end{aligned}$$

(A1)

It is obvious that $A_{1}>0$ and $A_{3}>0$ when the conditions in Theorem 7 are satisfied. By the Routh-Hurwitz criterion, it follows that $E_{xyz}$ is locally asymptotically stable if $A_{1}A_{2}>A_{3}$.

For the global asymptotic stability of $E_{xyz}$, consider the positive definite Lyapunov function

$$\begin{aligned} V(x,y,z)= & {} l_{1}\left( x-x^{*}-x^{*}\ln \frac{x}{x^{*}}\right) +\left( y-y^{*}-y^{*}\ln \frac{y}{y^{*}}\right) \\{} & {} \quad +l_{2}\left( z-z^{*}-z^{*}\ln \frac{z}{z^{*}}\right) , \end{aligned}$$

where $l_{1}$ and $l_{2}$ are positive constants to be determined later.

By calculating the derivative of V(x, y, z) along the solutions of (1), we obtain

$$\begin{aligned} \frac{\textrm{d}V}{\textrm{d}t}= & {} l_{1}(x-x^{*})\left[ r_{1}\left( 1-\frac{x}{K_{1}}\right) -\frac{m_{1}y}{a_{1}+x}-r_{1}\left( 1-\frac{x^{*}}{K_{1}}\right) +\frac{m_{1}y^{*}}{a_{1}+x^{*}}\right] \\{} & {} \quad \displaystyle +(y-y^{*})\left[ r_{2}\left( 1-\frac{y}{K_{2}}\right) +\frac{e_{1}m_{1}x}{a_{1}+x} \right. \nonumber \\{} & {} \quad \left. -\frac{m_{2}z}{a_{2}+y} -r_{2}\left( 1-\frac{y^{*}}{K_{2}}\right) -\frac{e_{1}m_{1}x^{*}}{a_{1}+x^{*}}+\frac{m_{2}z^{*}}{a_{2}+y^{*}}\right] \\{} & {} \quad \displaystyle +l_{2}(z-z^{*})\left( \frac{e_{2}m_{2}y}{a_{2}+y}-\frac{e_{2}m_{2}y^{*}}{a_{2}+y^{*}}\right) . \end{aligned}$$

After simplification,

$$\begin{aligned} \frac{\textrm{d}V}{\textrm{d}t}= & {} -l_{1}\left[ \frac{r_{1}}{K_{1}}-\frac{m_{1}y^{*}}{(a_{1}+x)(a_{1}+x^{*})}\right] (x-x^{*})^{2}\nonumber \\{} & {} \quad -\left[ \frac{r_{2}}{K_{2}}-\frac{m_{2}z^{*}}{(a_{2}+y)(a_{2}+y^{*})}\right] (y-y^{*})^{2}\nonumber \\{} & {} \quad \displaystyle +\left[ \frac{e_{1}m_{1}a_{1}}{(a_{1}+x)(a_{1}+x^{*})}-\frac{l_{1}m_{1}a_{1}}{(a_{1}+x)(a_{1}+x^{*})}\right. \nonumber \\{} & {} \quad \left. -\frac{l_{1}m_{1}x^{*}}{(a_{1}+x)(a_{1}+x^{*})}\right] (x-x^{*})(y-y^{*})\nonumber \\{} & {} \quad \displaystyle +\left[ \frac{l_{2}e_{2}m_{2}a_{2}}{(a_{2}+y)(a_{2}+y^{*})}\right. \nonumber \\{} & {} \quad \left. -\frac{m_{2}a_{2}}{(a_{2}+y)(a_{2}+y^{*})}-\frac{m_{2}y^{*}}{(a_{2}+y)(a_{2}+y^{*})}\right] (y-y^{*})(z-z^{*}). \end{aligned}$$

(A2)

Set

$$\begin{aligned} l_{1}=\frac{e_{1}a_{1}}{a_{1}+x^{*}},~~l_{2}=\frac{a_{2}+y^{*}}{e_{2}a_{2}} \end{aligned}$$

(A3)

such that the right hand side of (A2) can be written as $-X^{\textrm{T}}DX$, where $X^{\textrm{T}}=(x-x^{*},y-y^{*},z-z^{*})$ and the symmetric matrix D is given by

$$\begin{aligned} \begin{aligned} D=\left( \begin{array}{cccccc} \frac{l_1r_{1}}{K_{1}}-\frac{l_1m_{1}y^{*}}{(a_{1}+x)(a_{1}+x^{*})}&{}{}0&{}{}0\\ 0&{}{}\frac{r_{2}}{K_{2}}-\frac{m_{2}z^{*}}{(a_{2}+y)(a_{2}+y^{*})}&{}{}0\\ 0&{}{}0&{}{}0\\ \end{array} \right) . \end{aligned} \end{aligned}$$

Then ${\textrm{d}V}/{\textrm{d}t}<0$ if and only if the symmetric matrix D is positive definite, which is equivalent to (12). Therefore, $E_{xyz}$ is globally asymptotically stable. The proof is complete. $\square$

Appendix B: Proofs for Optimal Harvesting

1.1 Proof of Theorem 8

Proof

The bionomic equilibrium solves the system of algebraic equations

$$\begin{aligned} \left\{ \begin{aligned}&r_{1}x\left( 1-\frac{x}{K_{1}}\right) -\frac{m_{1}xy}{a_{1}+x}=0,\\&r_{2}y\left( 1-\frac{y}{K_{2}}\right) +\frac{e_{1}m_{1}xy}{a_{1}+x}-\frac{m_{2}yz}{a_{2}+y}-q_1Ey=0,\\&\frac{e_{2}m_{2}yz}{a_{2}+y}-dz-q_2Ez=0,\\&(p_1q_1y+p_2q_2z-c)E=0. \end{aligned} \right. \end{aligned}$$

(B4)

Define

$$\begin{aligned} \bar{\Omega }= & {} \Big \{(x,y,z,E)\mid 0\le x\le K_1,~\frac{a_2 d}{e_2m_2-d}\le y\le \frac{c}{p_1q_1},\\{} & {} \quad \displaystyle ~~~~0\le z\le \frac{c}{p_2q_2},~0\le E \le \frac{e_2m_2c}{q_2(c+a_2p_1q_1)}\Big \}. \end{aligned}$$

Obviously, $\bar{\Omega }$ is a positive invariant set of (B4). From the first equation, one has

$$\begin{aligned} \bar{y}=\frac{r_{1}}{m_{1}}\left( 1-\frac{\bar{x}}{K_{1}}\right) (a_{1}+\bar{x}). \end{aligned}$$

Substituting $\bar{y}$ into the third equation produces

$$\begin{aligned} \bar{E}=\frac{1}{q_2}\frac{e_{2}m_{2}\frac{r_{1}}{m_{1}}\left( 1-\frac{\bar{x}}{K_{1}}\right) (a_{1}+\bar{x})}{a_{2}+\frac{r_{1}}{m_{1}}\left( 1-\frac{\bar{x}}{K_{1}}\right) (a_{1}+\bar{x})}-\frac{d}{q_2}, \end{aligned}$$

where $\bar{E}>0$ since $d<\frac{e_{2}m_{2}\bar{y}}{a_{2}+\bar{y}}$. Similarly, we have

$$\begin{aligned} \bar{z}=\frac{c}{p_2q_2}-\frac{p_1q_1r_{1}}{p_2q_2m_{1}}\left( 1-\frac{\bar{x}}{K_{1}}\right) (a_{1}+\bar{x}), \end{aligned}$$

where $\bar{z}>0$ since $c>p_1q_1\bar{y}$. According to the second equation of (B4), we can see $\bar{x}$ is a positive root of the following equation

$$\begin{aligned}{} & {} r_{2}\left[ 1-\frac{r_{1}}{m_{1}K_{2}}\left( 1-\frac{x}{K_{1}}\right) (a_{1}+x)\right] +\frac{e_{1}m_{1}x}{a_{1}+x}\\{} & {} \qquad -\frac{m_{2}\left[ \frac{c}{p_2q_2}-\frac{p_1q_1r_{1}}{p_2q_2m_{1}}\left( 1-\frac{x}{K_{1}}\right) (a_{1}+x)\right] }{a_{2}+\frac{r_{1}}{m_{1}}\left( 1-\frac{x}{K_{1}}\right) (a_{1}+x)}\\{} & {} \qquad -\frac{q_1}{q_2}\frac{e_{2}m_{2}\frac{r_{1}}{m_{1}}\left( 1-\frac{x}{K_{1}}\right) (a_{1}+x)}{a_{2}+\frac{r_{1}}{m_{1}}\left( 1-\frac{x}{K_{1}}\right) (a_{1}+x)}+\frac{dq_1}{q_2}=0, \end{aligned}$$

which can be simplified by

$$\begin{aligned} \begin{aligned} -\frac{r_{1}^{2}r_{2}}{m_{1}^{2}K_{1}^{2}K_{2}}x^{5}+\Theta _{1} x^{4}+\Theta _{2}x^{3}+\Theta _{3}x^{2}+\Theta _{4}x+\Theta _{0}=0, \end{aligned} \end{aligned}$$

(B5)

where

$$\begin{aligned} \Theta _{0}=a_1\left( a_{2}+\frac{a_{1}r_{1}}{m_{1}}\right) \left( r_{2}-\frac{a_{1}r_{1}r_{2}}{m_{1}K_{2}}+\frac{dq_1}{q_2}\right) +\frac{a_{1}r_{1}m_{2}q_1}{m_{1}q_2}\left( \frac{p_1}{p_2}-e_{2}-\frac{m_{1}c}{a_{1}r_{1}q_1p_2}\right) \end{aligned}$$

must be a positive constant under the condition $r_{1}<\frac{m_{1}K_{2}}{a_{1}}$ and $\frac{p_1}{p_2}>e_{2}+\frac{m_{1}c}{a_{1}r_{1}q_1p_2}$. From Lemma 2, it follows that there exists at least one positive root $\bar{x}$. Therefore, the positive bionomic equilibrium exists if (13) is satisfied. The proof is complete. $\square$

1.2 Proof of Theorem 9

Proof

By Pontryagin’s maximum principle (1987), the adjoint equations read

$$\begin{aligned} \frac{\textrm{d}\chi _{1}}{\textrm{d}t}=-\frac{\partial \mathcal {F}}{\partial x},~~\frac{\textrm{d}\chi _{2}}{\textrm{d}t}=-\frac{\partial \mathcal {F}}{\partial y},~~\frac{\textrm{d}\chi _{3}}{\textrm{d}t}=-\frac{\partial \mathcal {F}}{\partial z}. \end{aligned}$$

Consider the equilibrium of (B4), then the adjoint equations reduce to

$$\begin{aligned}{} & {} \frac{\textrm{d}\chi _{1}}{\textrm{d}t}=\left( \frac{r_{1}}{K_{1}}x-\frac{m_{1}xy}{(a_{1}+x)^{2}}\right) \chi _{1}-\frac{e_{1}m_{1}a_{1}y}{(a_{1}+x)^{2}}\chi _{2}, \end{aligned}$$

(B6)

$$\begin{aligned}{} & {} \quad \frac{\textrm{d}\chi _{2}}{\textrm{d}t}=\frac{m_{1}x}{a_{1}+x}\chi _{1}\nonumber \\{} & {} +\left( \frac{r_{2}}{K_{2}}y-\frac{m_{2}yz}{(a_{2}+y)^{2}}\right) \chi _{2}-\frac{e_{2}m_{2}a_{2}z}{(a_{2}+y)^{2}}\chi _{3}\nonumber \\{} & {} \quad \displaystyle -\left( \xi _{1}\textrm{e}^{-\delta _{v}t}+\xi _{2}\textrm{e}^{-\delta _{u}t}\right) p_1q_1E, \end{aligned}$$

(B7)

$$\begin{aligned}{} & {} \quad \frac{\textrm{d}\chi _{3}}{\textrm{d}t}=\frac{m_{2}y}{a_{2}+y}\chi _{2}-\left( \xi _{1}\textrm{e}^{-\delta _{v}t}+\xi _{2}\textrm{e}^{-\delta _{u}t}\right) p_2q_2E. \end{aligned}$$

(B8)

Then

$$\begin{aligned}{} & {} \chi _{2}^{''}-\left( \frac{r_{1}}{K_{1}}x-\frac{m_{1}xy}{(a_{1}+x)^{2}}\right) \chi _{2}^{'}\nonumber \\{} & {} ~=-\left( \frac{r_{1}}{K_{1}}x-\frac{m_{1}xy}{(a_{1}+x)^{2}}\right) \left( \frac{r_{2}}{K_{2}}y-\frac{m_{2}yz}{(a_{2}+y)^{2}}\right) \chi _{2}\nonumber \\{} & {} \quad +\left( \frac{r_{1}}{K_{1}}x-\frac{m_{1}xy}{(a_{1}+x)^{2}}\right) \frac{e_{2}m_{2}a_{2}z}{(a_{2}+y)^{2}}\chi _{3}\nonumber \\{} & {} \quad \displaystyle +\left( \frac{r_{1}}{K_{1}}x-\frac{m_{1}xy}{(a_{1}+x)^{2}}\right) \left( \xi _{1}\textrm{e}^{-\delta _{v}t}+\xi _{2}\textrm{e}^{-\delta _{u}t}\right) p_1q_1E -\frac{m_{1}x}{a_{1}+x}\frac{e_{1}m_{1}a_{1}y}{(a_{1}+x)^{2}}\chi _{2}\nonumber \\{} & {} \quad \displaystyle {+}\left( \frac{r_{2}}{K_{2}}y{-}\frac{m_{2}yz}{(a_{2}{+}y)^{2}}\right) \chi _{2}^{'} {-}\frac{e_{2}m_{2}a_{2}z}{(a_{2}{+}y)^{2}}\left( \frac{m_{2}y}{a_{2}{+}y}\chi _{2}{-}\left( \xi _{1}\textrm{e}^{-\delta _{v}t}+\xi _{2}\textrm{e}^{-\delta _{u}t}\right) p_2q_2E\right) .\nonumber \\ \end{aligned}$$

(B9)

Combining (17) with (B9), one has

$$\begin{aligned} \chi _{2}^{''}+\mathcal {Q}_{1}\chi _{2}^{'}+\mathcal {Q}_{2}\chi _{2}=\mathcal {M}\textrm{e}^{-\delta _{v}t}+\mathcal {N}\textrm{e}^{-\delta _{u}t}, \end{aligned}$$

(B10)

where

$$\begin{aligned} \mathcal {Q}_{1}= & {} -\left( \frac{r_{1}}{K_{1}}x-\frac{m_{1}xy}{(a_{1}+x)^{2}}\right) -\left( \frac{r_{2}}{K_{2}}y-\frac{m_{2}yz}{(a_{2}+y)^{2}}\right) , \\ \mathcal {Q}_{2}= & {} \left( \frac{r_{1}}{K_{1}}x-\frac{m_{1}xy}{(a_{1}+x)^{2}}\right) \left( \frac{r_{2}}{K_{2}}y-\frac{m_{2}yz}{(a_{2}+y)^{2}}\right) \\{} & {} \quad +\left( \frac{r_{1}}{K_{1}}x-\frac{m_{1}xy}{(a_{1}+x)^{2}}\right) \frac{e_{2}m_{2}a_{2}z}{(a_{2}+y)^{2}}\frac{q_1y}{q_2z}\\{} & {} \quad \displaystyle +\frac{m_{1}x}{a_{1}+x}\frac{e_{1}m_{1}a_{1}y}{(a_{1}+x)^{2}}+\frac{e_{2}m_{2}a_{2}z}{(a_{2}+y)^{2}}\frac{m_{2}y}{a_{2}+y} \\ \mathcal {M}= & {} \left( \frac{r_{1}}{K_{1}}x-\frac{m_{1}xy}{(a_{1}+x)^{2}}\right) \frac{e_{2}m_{2}a_{2}z}{(a_{2}+y)^{2}}\frac{\xi _{1}(p_1q_1y+p_2q_2z-c)}{q_2z}\\{} & {} \quad \displaystyle +\left( \frac{r_{1}}{K_{1}}x-\frac{m_{1}xy}{(a_{1}+x)^{2}}\right) \xi _{1}p_1q_1E+\frac{e_{2}m_{2}a_{2}z}{(a_{2}+y)^{2}}\xi _{1}p_2q_2E, \\ \mathcal {N}= & {} \left( \frac{r_{1}}{K_{1}}x-\frac{m_{1}xy}{(a_{1}+x)^{2}}\right) \frac{e_{2}m_{2}a_{2}z}{(a_{2}+y)^{2}}\frac{\xi _{2}(p_1q_1y+p_2q_2z-c)}{q_2z}\\{} & {} \quad \displaystyle +\left( \frac{r_{1}}{K_{1}}x-\frac{m_{1}xy}{(a_{1}+x)^{2}}\right) \xi _{2}p_1q_1E+\frac{e_{2}m_{2}a_{2}z}{(a_{2}+y)^{2}}\xi _{2}p_2q_2E. \end{aligned}$$

This leads to the solution

$$\begin{aligned} \chi _{2}=\Xi _{1}\textrm{e}^{\triangle _{1}t}+\Xi _{2}\textrm{e}^{\triangle _{2}t}+\frac{\mathcal {M}}{\delta _{v}^{2}-\mathcal {Q}_{1}\delta _{v}+\mathcal {Q}_{2}}\textrm{e}^{-\delta _{v}t}+\frac{\mathcal {N}}{\delta _{u}^{2}-\mathcal {Q}_{1}\delta _{u}+\mathcal {Q}_{2}}\textrm{e}^{-\delta _{u}t}, \end{aligned}$$

where $\Xi _{1}$ and $\Xi _{2}$ are constants of integration, $\triangle _{1}$ and $\triangle _{2}$ are the roots of the auxiliary equation $\triangle ^{2}+\mathcal {Q}_{1}\triangle +\mathcal {Q}_{2}=0$. Imposing the transverse condition $\chi _{2}(t)\rightarrow 0$ as $t\rightarrow \infty$ leads to $\Xi _{1}=\Xi _{2}=0$. Then,

$$\begin{aligned} \chi _{2}=\frac{\mathcal {M}}{\delta _{v}^{2}-\mathcal {Q}_{1}\delta _{v}+\mathcal {Q}_{2}}\textrm{e}^{-\delta _{v}t}+\frac{\mathcal {N}}{\delta _{u}^{2}-\mathcal {Q}_{1}\delta _{u}+\mathcal {Q}_{2}}\textrm{e}^{-\delta _{u}t}. \end{aligned}$$

Substituting $\chi _2$ into (17) reaches

$$\begin{aligned} \chi _{3}= & {} \left( \frac{p_1q_1y+p_2q_2z-c}{q_2z}\xi _{1}-\frac{q_1y}{q_2z}\frac{\mathcal {M}}{\delta _{v}^{2}-\mathcal {Q}_{1}\delta _{v}+\mathcal {Q}_{2}}\right) \textrm{e}^{-\delta _{v}t}\\{} & {} \quad \displaystyle +\left( \frac{p_1q_1y+p_2q_2z-c}{q_2z}\xi _{2}-\frac{q_1y}{q_2z}\frac{\mathcal {N}}{\delta _{u}^{2}-\mathcal {Q}_{1}\delta _{u}+\mathcal {Q}_{2}}\right) \textrm{e}^{-\delta _{u}t}. \end{aligned}$$

Similarly,

$$\begin{aligned} \chi _{1}^{''}+\tilde{\mathcal {Q}}_{1}\chi _{1}^{'}+\tilde{\mathcal {Q}}_{2}\chi _{1}=\tilde{\mathcal {M}}\textrm{e}^{-\delta _{v}t}+\tilde{\mathcal {N}}\textrm{e}^{-\delta _{u}t}, \end{aligned}$$

(B11)

where

$$\begin{aligned} \tilde{\mathcal {Q}}_{1}= & {} \frac{m_{1}xy}{(a_{1}+x)^{2}}-\frac{r_{1}}{K_{1}}x,~~~\tilde{\mathcal {Q}}_{2}=\frac{e_{1}m_{1}^{2}a_{1}xy}{(a_{1}+x)^{3}}, \\ \tilde{\mathcal {M}}= & {} -\frac{e_{1}m_{1}a_{1}y}{(a_{1}+x)^{2}}\left( \frac{r_{2}}{K_{2}}y-\frac{m_{2}yz}{(a_{2}+y)^{2}}\right) \frac{\mathcal {M}}{\delta _{v}^{2}-\mathcal {Q}_{1}\delta _{v}+\mathcal {Q}_{2}}+\frac{e_{1}m_{1}a_{1}y}{(a_{1}+x)^{2}}p_1q_1E\xi _{1}\\{} & {} \displaystyle +\frac{e_{1}m_{1}a_{1}y}{(a_{1}+x)^{2}}\frac{e_{2}m_{2}a_{2}z}{(a_{2}+y)^{2}}\left( \frac{p_1q_1y+p_2q_2z-c}{q_2z}\xi _{1}-\frac{q_1y}{q_2z}\frac{\mathcal {M}}{\delta _{v}^{2}-\mathcal {Q}_{1}\delta _{v}+\mathcal {Q}_{2}}\right) , \\ \tilde{\mathcal {N}}= & {} -\frac{e_{1}m_{1}a_{1}y}{(a_{1}+x)^{2}}\left( \frac{r_{2}}{K_{2}}y-\frac{m_{2}yz}{(a_{2}+y)^{2}}\right) \frac{\mathcal {N}}{\delta _{u}^{2}-\mathcal {Q}_{1}\delta _{u}+\mathcal {Q}_{2}}+\frac{e_{1}m_{1}a_{1}y}{(a_{1}+x)^{2}}p_1q_1E\xi _{2}\\{} & {} \displaystyle +\frac{e_{1}m_{1}a_{1}y}{(a_{1}+x)^{2}}\frac{e_{2}m_{2}a_{2}z}{(a_{2}+y)^{2}}\left( \frac{p_1q_1y+p_2q_2z-c}{q_2z}\xi _{2}-\frac{q_1y}{q_2z}\frac{\mathcal {N}}{\delta _{u}^{2}-\mathcal {Q}_{1}\delta _{u}+\mathcal {Q}_{2}}\right) . \end{aligned}$$

$\chi _{1}$ can be simplified as

$$\begin{aligned} \chi _{1}=\frac{\tilde{\mathcal {M}}}{\delta _{v}^{2}-\tilde{\mathcal {Q}}_{1}\delta _{v}+\tilde{\mathcal {Q}}_{2}}\textrm{e}^{-\delta _{v}t}+\frac{\tilde{\mathcal {N}}}{\delta _{u}^{2}-\tilde{\mathcal {Q}}_{1}\delta _{u}+\tilde{\mathcal {Q}}_{2}}\textrm{e}^{-\delta _{u}t}. \end{aligned}$$

It is not difficult to find that the shadow prices $\chi _{1}(t)\textrm{e}^{\delta _{u}t}$, $\chi _{2}(t)\textrm{e}^{\delta _{u}t}$, and $\chi _{3}(t)\textrm{e}^{\delta _{u}t}$ are bounded and constant over time in optimal equilibrium when they satisfy the transversality condition at $\infty$ (Sadhukhan et al 2010).

Substituting the values of $\chi _{2},\chi _{3}$ into (B8), one obtains

$$\begin{aligned} E=&{} \Bigg \{\left[ \delta _{v}\left( \frac{p_1q_1y+p_2q_2z-c}{q_2z}\xi _{1}-\frac{q_1y}{q_2z}\frac{\mathcal {M}}{\delta _{v}^{2}-\mathcal {Q}_{1}\delta _{v}+\mathcal {Q}_{2}}\right) +\frac{m_{2}y}{a_{2}+y}\frac{\mathcal {M}}{\delta _{v}^{2}-\mathcal {Q}_{1}\delta _{v}+\mathcal {Q}_{2}}\right] \text {e}^{-\delta _{v}t}\nonumber \\{}&{} \quad \displaystyle +\left[ \delta _{u}\left( \frac{p_1q_1y+p_2q_2z-c}{q_2z}\xi _{2}-\frac{q_1y}{q_2z}\frac{\mathcal {N}}{\delta _{u}^{2}-\mathcal {Q}_{1}\delta _{u}+\mathcal {Q}_{2}}\right) +\frac{m_{2}y}{a_{2}+y}\frac{\mathcal {N}}{\delta _{u}^{2}-\mathcal {Q}_{1}\delta _{u}+\mathcal {Q}_{2}}\right] \text {e}^{-\delta _{u}t}\Bigg \}\nonumber \\{}&{} \quad \displaystyle 1/\left[ \left( \xi _{1}\text {e}^{-\delta _{v}t}+\xi _{2}\text {e}^{-\delta _{u}t}\right) p_2q_2\right] . \end{aligned}$$

(B12)

The optimal equilibrium solution $(x^{**},y^{**},z^{**},E^{**})$ of the control problem can be obtained by solving steady state equations together with (B12). Moreover, the maximum net present revenue is

$$\begin{aligned} \mathcal {J}(y^{**},z^{**},E^{**})= & {} \int _{0}^{\infty }\textrm{e}^{-\tilde{\delta }t}(p_1q_1y^{**}+p_2q_2z^{**}-c)E^{**}\textrm{d}t\\= & {} \frac{(p_1q_1y^{**}+p_2q_2z^{**}-c)E^{**}}{\tilde{\delta }}. \end{aligned}$$

The proof is complete. $\square$

Appendix C: Proofs for Stochastic System

1.1 Proof of Theorem 10

Proof

Since (2) has a unique local solution $(x(t),y(t),z(t))\in \mathbb {R}^{3}_{+}$ on $[0,\tau _{e})$ a.s., where $\tau _{e}$ is the explosion time. In order to show that this solution is global, i.e., $\tau _{e}=\infty$ a.s., let $n_{0}\ge 0$ be sufficiently large such that $x_{0}$, $y_{0}$, and $z_{0}$ are lying within the interval $[{1}/{n_{0}},n_{0}]$. For each integer $n\ge n_{0}$, define the stopping time as

$$\begin{aligned} \tau _{n}=\inf \left\{ t\in [0,\tau _{e}):\min \{x(t),y(t),z(t)\}\le \frac{1}{n}~\text {or}~\max \{x(t),y(t),z(t)\}\ge n\right\} . \end{aligned}$$

Obviously, $\tau _{n}$ is increasing as $n\rightarrow \infty$. Set $\tau _{\infty }=\lim _{n\rightarrow \infty } \tau _{n}$, whence $\tau _{\infty }\le \tau _{e}$ a.s. Hence, it only needs to prove that $\tau _{\infty }=\infty$ a.s. It can be proved by contradiction arguments. Assume that the statement is not true, then there exist a pair of constants $T>0$ and $\kappa \in (0,1)$ such that $\mathbb {P}\left\{ \tau _{\infty }\le T\right\} >\kappa .$ Whence there exists an integer $n_{1}\ge n_{0}$ such that $\mathbb {P}\left\{ \tau _{n}\le T\right\} \ge \kappa ,~n\ge n_{1}.$ Define a $C^{2}$-function V: $\mathbb {R}^{3}_{+}\rightarrow \mathbb {R}_{+}$ by

$$\begin{aligned} V(x,y,z)= & {} x^{\theta }-\ln x-\frac{1}{\theta }\left( 1-\ln \frac{1}{\theta }\right) +y^{\theta }-\ln y-\frac{1}{\theta }\left( 1-\ln \frac{1}{\theta }\right) \\{} & {} \quad +z^{\theta }-\ln z-\frac{1}{\theta }\left( 1-\ln \frac{1}{\theta }\right) \end{aligned}$$

with $0<\theta <1$. The nonnegativity of V follows from $u^{\theta }-\ln u-(1-\ln (1/\theta ))/\theta \ge 0,~u\ge 0.$ By applying Itô’s formula, one has

$$\begin{aligned} \begin{array}{l} \mathcal {L}(x^{\theta })=\theta r_{1}x^{\theta }\left( 1-\frac{x}{K_{1}}\right) -\frac{m_{1}\theta x^{\theta }y}{a_{1}+x}+ \frac{1}{2}\theta (\theta -1)x^{\theta }(\sigma _{11}+\sigma _{12}x)^{2},\\ \mathcal {L}(y^{\theta })=\theta r_{2}y^{\theta }\left( 1-\frac{y}{K_{2}}\right) +\frac{\theta e_{1}m_{1}xy^{\theta }}{a_{1}+x} -\frac{\theta m_{2}y^{\theta }z}{a_{2}+y}\\ -\theta h_{1}y^{\theta }+\frac{1}{2}\theta (\theta -1)y^{\theta }(\sigma _{21}+\sigma _{22}y)^{2},\\ \mathcal {L}(z^{\theta })=\frac{\theta e_{2}m_{2}z^{\theta }y}{a_{2}+y}-\theta dz^{\theta }-\theta h_{2}z^{\theta }+\frac{1}{2}\theta (\theta -1)z^{\theta }(\sigma _{31}+\sigma _{32}z)^{2},\\ \mathcal {L}(\ln x)=r_{1}\left( 1-\frac{x}{K_{1}}\right) -\frac{m_{1}y}{a_{1}+x}-\frac{1}{2}(\sigma _{11}+\sigma _{12}x)^{2},\\ \mathcal {L}(\ln y)=r_{2}\left( 1-\frac{y}{K_{2}}\right) +\frac{e_{1}m_{1}x}{a_{1}+x}-\frac{m_{2}z}{a_{2}+y}-h_{1} -\frac{1}{2}(\sigma _{21}+\sigma _{22}y)^{2},\\ \mathcal {L}(\ln z)=\frac{e_{2}m_{2}y}{a_{2}+y}-d-h_{2}-\frac{1}{2}(\sigma _{31}+\sigma _{32}z)^{2}. \end{array} \end{aligned}$$

Then, by Lemma 3,

$$\begin{aligned} \mathcal {L}V= & {} \theta r_{1}x^{\theta }\left( 1-\frac{x}{K_{1}}\right) -\frac{m_{1}\theta x^{\theta }y}{a_{1}+x}+ \frac{\theta (\theta -1)}{2}x^{\theta }(\sigma _{11}+\sigma _{12}x)^{2} -r_{1}\left( 1-\frac{x}{K_{1}}\right) \\ {}{} & {} +\frac{m_{1}y}{a_{1}+x}+\frac{1}{2}(\sigma _{11}+\sigma _{12}x)^{2} +\theta r_{2}y^{\theta }\left( 1-\frac{y}{K_{2}}\right) +\frac{\theta e_{1}m_{1}xy^{\theta }}{a_{1}+x} -\frac{\theta m_{2}y^{\theta }z}{a_{2}+y}\\ {}{} & {} -\theta h_{1}y^{\theta }+\frac{\theta (\theta -1)}{2}y^{\theta }(\sigma _{21}+\sigma _{22}y)^{2} -r_{2}\left( 1-\frac{y}{K_{2}}\right) -\frac{e_{1}m_{1}x}{a_{1}+x}+\frac{m_{2}z}{a_{2}+y}\\ {}{} & {} +h_{1}+\frac{1}{2}(\sigma _{21}+\sigma _{22}y)^{2} +\frac{\theta e_{2}m_{2}z^{\theta }y}{a_{2}+y}-\theta dz^{\theta }-\theta h_{2}z^{\theta }\\ {}{} & {} +\frac{\theta (\theta -1)}{2}z^{\theta }(\sigma _{31}+\sigma _{32}z)^{2} -\frac{e_{2}m_{2}y}{a_{2}+y}+d+h_{2}+\frac{1}{2}(\sigma _{31}+\sigma _{32}z)^{2}\\\le & {} -\frac{\theta (1-\theta )}{2}\sigma _{12}^{2}x^{\theta +2}+\frac{1}{2}\sigma _{12}^{2}x^{2}+\left( \frac{r_{1}}{K_{1}}+\sigma _{11}\sigma _{12}\right) x+\theta r_{1}x^{\theta }\\{} & {} -\frac{\theta (1-\theta )}{2})\sigma _{22}^{2}y^{\theta +2}+\frac{1}{2}\sigma _{22}^{2}y^{2}+\left( \frac{r_{2}}{K_{2}}+\frac{m_{1}}{a_{1}} +\sigma _{21}\sigma _{22}\right) y+(\theta r_{2}+\theta e_{1}m_{1})y^{\theta }\\{} & {} -\frac{\theta (1-\theta )}{2}\sigma _{32}^{2}z^{\theta +2}+\frac{1}{2}\sigma _{32}^{2}z^{2}+\left( \frac{m_{2}}{a_{2}}+\sigma _{31}\sigma _{32}\right) z+\theta e_{2}m_{2}z^{\theta }\\{} & {} +d+h_{1}+h_{2}+\frac{1}{2}\sigma _{11}^{2}+\frac{1}{2}\sigma _{21}^{2}+\frac{1}{2}\sigma _{31}^{2}\\\le & {} \rho _1+\rho _2+\rho _3+d+h_{1}+h_{2}+\frac{1}{2}\sigma _{11}^{2}+\frac{1}{2}\sigma _{21}^{2}+\frac{1}{2}\sigma _{31}^{2}:= \rho , \end{aligned}$$

where

$$\begin{aligned} \begin{array}{l} \rho _1=\sup \limits _{(x,y,z)\in \mathbb {R}^{3}_{+}}\left\{ -\frac{\theta (1-\theta )}{2}\sigma _{12}^{2}x^{\theta +2}+\frac{\sigma _{12}^{2}}{2}x^{2}+\left( \frac{r_{1}}{K_{1}}+\sigma _{11}\sigma _{12}\right) x+\theta r_{1}x^{\theta }\right\} ,\\ \rho _2{=}\sup \limits _{(x,y,z)\in \mathbb {R}^{3}_{+}}\left\{ -\frac{\theta (1{-}\theta )}{2}\sigma _{22}^{2}y^{\theta {+}2}{+}\frac{\sigma _{22}^{2}}{2}y^{2}{+}\left( \frac{r_{2}}{K_{2}}{+}\frac{m_{1}}{a_{1}} {+}\sigma _{21}\sigma _{22}\right) y{+}(\theta r_{2}{+}\theta e_{1}m_{1})y^{\theta }\right\} ,\\ \rho _3=\sup \limits _{(x,y,z)\in \mathbb {R}^{3}_{+}}\left\{ -\frac{\theta (1-\theta )}{2}\sigma _{32}^{2}z^{\theta +2}+\frac{\sigma _{32}^{2}}{2}z^{2}+\left( \frac{m_{2}}{a_{2}}+\sigma _{31}\sigma _{32}\right) z+\theta e_{2}m_{2}z^{\theta }\right\} , \end{array} \end{aligned}$$

and $\rho$ is a positive constant being independent of x, y, z, and t. Consequently,

$$\begin{aligned} \textrm{d}V\le & {} \rho \textrm{d}t+(\theta x^{\theta }-1)(\sigma _{11}+\sigma _{12}x)\textrm{d}B_{1}(t)\\{} & {} +(\theta y^{\theta }-1)(\sigma _{21}+\sigma _{22}y)\textrm{d}B_{2}(t)+(\theta z^{\theta }-1)(\sigma _{31}+\sigma _{32}z)\textrm{d}B_{3}(t). \end{aligned}$$

Integrating it from 0 to $\tau _{n}\wedge T$ and taking expectation produces

$$\begin{aligned} \mathbb {E}(V(x(\tau _{n}\wedge T),y(\tau _{n}\wedge T),z(\tau _{n}\wedge T)))\le V(x_{0},y_{0},z_{0})+\rho T. \end{aligned}$$

Set $\Omega _{n}=\left\{ \omega \in \Omega : \tau _{n}=\tau _{n}(\omega )\le T\right\}$, then $\mathbb {P}(\Omega _{n})\ge \kappa$, and, for any $\omega \in \Omega _{n}$, $x(\tau _{n},\omega )$, $y(\tau _{n},\omega )$ or $z(\tau _{n},\omega )$, equals either n or 1/n. Hence,

$$\begin{aligned} \begin{array}{rcl} V(x_{0},y_{0},z_{0})+\rho T&{}\ge &{} \mathbb {E}(I_{\Omega _{n}}V(x(\tau _{n}),y(\tau _{n}),z(\tau _{n}))\\ &{}\ge &{} \rho \min \left\{ n^{\theta }-\ln n-\frac{1}{\theta }\left( 1-\ln \frac{1}{\theta }\right) ,~ n^{-\theta }+\ln n-\frac{1}{\theta }\left( 1-\ln \frac{1}{\theta }\right) \right\} , \end{array} \end{aligned}$$

where $I_{\Omega _{n}}$ denotes the indicator function of $\Omega _{n}$. Let $n\rightarrow \infty$, then there is a contradiction

$$\begin{aligned} \infty >V(x_{0},y_{0},z_{0})+\rho T=\infty . \end{aligned}$$

Therefore, $\tau _{\infty }=\infty$. The proof is complete. $\square$

1.2 Proof of Theorem 11

Proof

Define

$$\begin{aligned} \tilde{V}(x,y,z)=(e_1e_2x+e_2y+z)^{\tilde{\theta }}. \end{aligned}$$

By the generalized Itô’s formula, one has

$$\begin{aligned} \textrm{d}\tilde{V}= & {} \mathcal {L}\tilde{V}\textrm{d}t+e_1e_2\tilde{\theta }(e_1e_2x+e_2y+z)^{\tilde{\theta }-1}x(\sigma _{11}+\sigma _{12}x)\text {d}B_{1}(t)\\{} & {} \displaystyle +e_2\tilde{\theta }(e_1e_2x+e_2y+z)^{\tilde{\theta }-1}y(\sigma _{21}+\sigma _{22}y)\text {d}B_{2}(t)\\{} & {} \displaystyle +\tilde{\theta }(e_1e_2x+e_2y+z)^{\tilde{\theta }-1}z(\sigma _{31}+\sigma _{32}z)\text {d}B_{3}(t), \end{aligned}$$

where

$$\begin{aligned} \mathcal {L}\tilde{V}= & {} \tilde{\theta }(e_1e_2x+e_2y+z)^{\tilde{\theta }-1}\left( e_1e_2r_1x\left( 1-\frac{x}{K_1}\right) +e_2r_2y\left( 1-\frac{y}{K_2}\right) \right. \\{} & {} \left. -e_2h_1y-(d+h_2)z\right) \\{} & {} \displaystyle -\frac{\tilde{\theta }(1-\tilde{\theta })}{2}(e_1e_2x+e_2y+z)^{\tilde{\theta }-2}\\{} & {} \displaystyle \times \left( [e_1e_2x(\sigma _{11}+\sigma _{12}x)]^2 +[e_2y(\sigma _{21}+\sigma _{22}y)]^2+[z(\sigma _{31}+\sigma _{32}z)]^2\right) . \end{aligned}$$

For any $0<\beta <d+h_2$, applying Itô’s formula to $\textrm{e}^{\beta t}\tilde{V}$ leads to

$$\begin{aligned}{} & {} \textrm{e}^{\beta t}(e_1e_2x(t)+e_2y(t)+z(t))^{\tilde{\theta }}\\{} & {} \quad =(e_1e_2x_0+e_2y_0+z_0)^{\tilde{\theta }}+\displaystyle \int \limits _{0}^{t}\textrm{e}^{\beta s}G(\tilde{X}(s))\textrm{d}s\\{} & {} \qquad +e_1e_2\tilde{\theta }\displaystyle \int \limits _{0}^{t}\textrm{e}^{\beta s}((e_1e_2x(s)+e_2y(s)+z(s))^{\tilde{\theta }-1}x(s)(\sigma _{11}+\sigma _{12}x(s)))\text {d}B_{1}(s)\\{} & {} \qquad +e_2\tilde{\theta }\displaystyle \int \limits _{0}^{t}\textrm{e}^{\beta s}((e_1e_2x(s)+e_2y(s)+z(s))^{\tilde{\theta }-1}y(s)(\sigma _{21}+\sigma _{22}y(s)))\text {d}B_{2}(s)\\{} & {} \qquad +\tilde{\theta }\displaystyle \int \limits _{0}^{t}\textrm{e}^{\beta s}((e_1e_2x(s)+e_2y(s)+z(s))^{\tilde{\theta }-1}z(s)(\sigma _{31}+\sigma _{32}z(s)))\text {d}B_{3}(s), \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} G(\tilde{X})&=\beta (e_1e_2x+e_2y+z)^{\tilde{\theta }}+\tilde{\theta }(e_1e_2x+e_2y+z)^{\tilde{\theta }-1}\\&\quad \displaystyle \times \left( e_1e_2r_1x\left( 1-\frac{x}{K_1}\right) +e_2r_2y\left( 1-\frac{y}{K_2}\right) -e_2h_1y-(d+h_2)z\right) \\&\quad \displaystyle -\frac{\tilde{\theta }(1-\tilde{\theta })}{2}(e_1e_2x+e_2y+z)^{\tilde{\theta }-2}\\&\quad \displaystyle \times ([e_1e_2x(\sigma _{11}+\sigma _{12}x)]^{2} +[e_2y(\sigma _{21}+\sigma _{22}y)]^{2}+[z(\sigma _{31}+\sigma _{32}z)]^{2}). \end{aligned} \end{aligned}$$

On the one hand,

$$\begin{aligned} \begin{aligned}&[e_1e_2x(\sigma _{11}+\sigma _{12}x)]^{2}+[e_2y(\sigma _{21}+\sigma _{22}y)]^{2}+[z(\sigma _{31}+\sigma _{32}z)]^{2}\\&~~~~~~~~\ge (e_1e_2\sigma _{12})^2x^4+(e_2\sigma _{22})^2y^4+\sigma _{32}^2z^4\\&~~~~~~~~\ge \min \{(e_1e_2\sigma _{12})^2,(e_2\sigma _{22})^2,\sigma _{32}^2\}(x^4+y^4+z^4); \end{aligned} \end{aligned}$$

on the other side,

$$\begin{aligned} \begin{aligned} (e_1e_2x+e_2y+z)^4&\le 3^3[(e_1e_2x)^4+(e_2y)^4+z^4]\\&\le 3^3\max \{(e_1e_2)^4,e_2^{4},1\}(x^4+y^4+z^4)\\&\le 3^3(x^4+y^4+z^4), \end{aligned} \end{aligned}$$

where the following H$\mathrm {\ddot{o}}$lder inequality has been used

$$\begin{aligned} \mid \sum _{i=1}^{k}p_{i}\mid ^{n}\le k^{n-1}\sum _{i=1}^{k}\mid p_{i}\mid ^{n}. \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned} G(\tilde{X})&\le \beta (e_1e_2x+e_2y+z)^{\tilde{\theta }}+\tilde{\theta }(e_1e_2x+e_2y+z)^{\tilde{\theta }-1}\\&\quad \displaystyle \times \Bigg (e_1e_2r_1x\left( 1-\frac{x}{K_1}\right) +e_2r_2y\left( 1-\frac{y}{K_2}\right) -e_2h_1y-(d+h_2)z\Bigg )\\&\quad \displaystyle -\frac{\tilde{\theta }(1-\tilde{\theta })}{54}(e_1e_2x+e_2y+z)^{\tilde{\theta }+2}\min \{(e_1e_2\sigma _{12})^2, (e_2\sigma _{22})^2, \sigma _{32}^2\}, \end{aligned} \end{aligned}$$

which is bounded, say by $\mathcal {G}^{*}(\tilde{\theta })$. Then

$$\begin{aligned} \begin{array}{l} \mathbb {E}\left[ \textrm{e}^{\beta (t\wedge \tau _{n})}(e_1e_2x((t\wedge \tau _{n}))+e_2y((t\wedge \tau _{n}))+z((t\wedge \tau _{n})))^{\tilde{\theta }}\right] \\ \quad \quad \quad \le (e_1e_2x_0+e_2y_0+z_0)^{\tilde{\theta }}+\mathcal {G}^{*}(\tilde{\theta })\displaystyle \int \limits _{0}^{t\wedge \tau _{n}}\textrm{e}^{\beta s}\textrm{d}s. \end{array} \end{aligned}$$

Letting $n\rightarrow \infty$ and then $t\rightarrow \infty$ yields

$$\begin{aligned} \begin{array}{l} \limsup \limits _{t\rightarrow \infty }\mathbb {E}\left[ (e_1e_2x(t)+e_2y(t)+z(t))^{\tilde{\theta }}\right] \\ \quad \quad \le \lim \limits _{t\rightarrow \infty }\textrm{e}^{-\beta t}\left( (e_1e_2x_0+e_2y_0+z_0)^{\tilde{\theta }}+\frac{\mathcal {G}^{*}(\tilde{\theta })\left( \textrm{e}^{\beta t}-1\right) }{\beta }\right) \\ \quad \quad =\frac{\mathcal {G}^{*}(\tilde{\theta })}{\beta }. \end{array} \end{aligned}$$

It follows that

$$\begin{aligned} \limsup _{t\rightarrow \infty }\mathbb {E}x^{\tilde{\theta }}\le \frac{\mathcal {G}^{*}(\tilde{\theta })}{\beta (e_1e_2)^{\tilde{\theta }}},~~ \limsup _{t\rightarrow \infty }\mathbb {E}y^{\tilde{\theta }}\le \frac{\mathcal {G}^{*}(\tilde{\theta })}{\beta e_2^{\tilde{\theta }}},~~ \limsup _{t\rightarrow \infty }\mathbb {E}z^{\tilde{\theta }}\le \frac{\mathcal {G}^{*}(\tilde{\theta })}{\beta }. \end{aligned}$$

Set

$$\begin{aligned} \mathcal {G}(\tilde{\theta })=\frac{\mathcal {G}^{*}(\tilde{\theta })}{\beta }\max \left\{ (e_1e_2)^{-\tilde{\theta }}, e_2^{-\tilde{\theta }},1\right\} , \end{aligned}$$

then

$$\begin{aligned} \limsup _{t\rightarrow \infty }\mathbb {E}x^{\tilde{\theta }}\le \mathcal {G}(\tilde{\theta }),~~~ \limsup _{t\rightarrow \infty }\mathbb {E}y^{\tilde{\theta }}\le \mathcal {G}(\tilde{\theta }),~~~ \limsup _{t\rightarrow \infty }\mathbb {E}z^{\tilde{\theta }}\le \mathcal {G}(\tilde{\theta }). \end{aligned}$$

The proof is complete. $\square$

1.3 Proof of Theorem 12

Proof

Note that, for any $(x_{0},y_{0},z_{0})\in \mathbb {R}^{3}_{+}$, the solution of (2) is positive, then

$$\begin{aligned} \text {d}x\le r_{1}x\left( 1-\frac{x}{K_{1}}\right) \text {d}t+x(\sigma _{11}+\sigma _{12}x)\text {d}B_{1}(t). \end{aligned}$$

Consider the following auxiliary one-dimensional stochastic differential equation

$$\begin{aligned} \textrm{d}X(t)=r_{1}X\left( 1-\frac{X}{K_{1}}\right) \text {d}t+X(\sigma _{11}+\sigma _{12}X)\text {d}B_{1}(t) \end{aligned}$$

(C13)

with an initial value $X(0)=x_{0}>0$. Set

$$\begin{aligned} f(x)=r_{1}x\left( 1-\frac{x}{K_{1}}\right) ,~~\sigma (x)=x(\sigma _{11}+\sigma _{12}x),~~x\in (0,\infty ). \end{aligned}$$

Direct calculation shows that

$$\begin{aligned} \int \frac{f(\upsilon )}{\sigma ^{2}(\upsilon )}\textrm{d}\upsilon =\frac{r_{1}}{\sigma _{11}^{2}}\ln \frac{\upsilon }{\sigma _{11}+\sigma _{12}\upsilon } +\frac{r_{1}(\sigma _{11}+K_{1}\sigma _{12})}{K_{1}\sigma _{11}\sigma _{12}(\sigma _{11}+\sigma _{12}\upsilon )}+Q. \end{aligned}$$

Then

$$\begin{aligned} \begin{array}{l} \displaystyle \int \limits _{0}^{\infty }\frac{1}{\sigma ^{2}(x)} \exp \left\{ \int _{1}^{x}\frac{2f(\upsilon )}{\sigma ^{2}(\upsilon )}\textrm{d}\upsilon \right\} \textrm{d}x\\ \quad \quad =\mathcal {C}\displaystyle \int \limits _{0}^{\infty }x^{\frac{2r_{1}}{\sigma _{11}^{2}}-2}(\sigma _{11}+\sigma _{12}x)^{-\frac{2r_{1}}{\sigma _{11}^{2}}-2} \exp \left\{ \frac{2r_{1}(\sigma _{11}+K_{1}\sigma _{12})}{K_{1}\sigma _{11}\sigma _{12}(\sigma _{11}+\sigma _{12}x)}\right\} \textrm{d}x, \end{array} \end{aligned}$$

where

$$\begin{aligned} \mathcal {C}=(\sigma _{11}+\sigma _{12})^{\frac{2r_{1}}{\sigma _{11}^{2}}} \exp \left\{ -\frac{2r_{1}(\sigma _{11}+K_{1}\sigma _{12})}{K_{1}\sigma _{11}\sigma _{12}(\sigma _{11}+\sigma _{12})}\right\} . \end{aligned}$$

Letting

$$\begin{aligned}{} & {} \tilde{\phi }(x)=x^{2(r_{1}/\sigma _{11}^{2}-1)}(\sigma _{11}+\sigma _{12}x)^{-2(r_{1}/\sigma _{11}^{2}+1)} \exp \left\{ \frac{2r_{1}(\sigma _{11}+K_{1}\sigma _{12})}{K_{1}\sigma _{11}\sigma _{12}(\sigma _{11}+\sigma _{12}x)}\right\} , \\{} & {} \quad \tilde{\phi }_{1}(x)=x^{2(r_{1}/\sigma _{11}^{2}-1)}\exp \left\{ \frac{2r_{1}(\sigma _{11}+K_{1}\sigma _{12})}{K_{1}\sigma _{11}\sigma _{12}(\sigma _{11}+\sigma _{12}x)}\right\} , \end{aligned}$$

then

$$\begin{aligned} \lim _{x\rightarrow 0^{+}}\frac{\tilde{\phi }(x)}{\tilde{\phi }_{1}(x)}=\sigma _{11}^{-2(r_1/\sigma _{11}^2+1)}>0, \end{aligned}$$

which implies that $\displaystyle \int \limits _{0}^{1}\tilde{\phi }(x)\textrm{d}x$ has the same convergence as $\displaystyle \int \limits _{0}^{1}\tilde{\phi }_{1}(x)\textrm{d}x$. Since

$$\begin{aligned} \displaystyle \int \limits _{0}^{1}\tilde{\phi }_{1}(x)\textrm{d}x\le \exp \left\{ \frac{2r_{1}(\sigma _{11}+K_{1}\sigma _{12})}{K_{1}\sigma _{11}^{2}\sigma _{12}}\right\} \displaystyle \int \limits _{0}^{1}x^{2(r_{1}/\sigma _{11}^{2}-1)}\textrm{d}x \end{aligned}$$

and $r_{1}>\sigma _{11}^{2}/2$, one has $\displaystyle \int \limits _{0}^{1}x^{2(r_{1}/\sigma _{11}^{2}-1)}\textrm{d}x<+\infty .$ On th

$$\begin{aligned} \tilde{\phi }_{2}(x)=x^{-2}\exp \left\{ \frac{2r_{1}(\sigma _{11}+K_{1}\sigma _{12})}{K_{1}\sigma _{11}\sigma _{12}(\sigma _{11}+\sigma _{12}x)}\right\} , \end{aligned}$$

then

$$\begin{aligned} \lim _{x\rightarrow +\infty }\frac{\tilde{\phi }(x)}{\tilde{\phi }_{2}(x)}=\sigma _{12}^{-2(r_1/\sigma _{11}^2+1)}>0, \end{aligned}$$

which implies that $\displaystyle \int \limits _{1}^{\infty }\tilde{\phi }(x)\textrm{d}x$ has the same convergence as $\displaystyle \int \limits _{1}^{\infty }\tilde{\phi }_{2}(x)\textrm{d}x$. Then

$$\begin{aligned} \displaystyle \int \limits _{1}^{\infty }\tilde{\phi }_{2}(x)\textrm{d}x\le \exp \left\{ \frac{2r_{1}(\sigma _{11}+K_{1}\sigma _{12})}{K_{1}\sigma _{11}\sigma _{12}(\sigma _{11}+\sigma _{12})}\right\} \displaystyle \int \limits _{1}^{\infty }\frac{1}{x^{2}}\textrm{d}x<+\infty . \end{aligned}$$

Whence

$$\begin{aligned} \displaystyle \int \limits _{0}^{\infty }\frac{1}{\sigma ^{2}(x)}\exp \left\{ \displaystyle \int \limits _{1}^{x}\frac{2f(\upsilon )}{\sigma ^{2}(\upsilon )} \textrm{d}\upsilon \right\} \textrm{d}x<\infty . \end{aligned}$$

Due to the ergodic property (Kutoyants and Kutojanc 2004), the invariant density of (C13) is

$$\begin{aligned} \varphi (x)=Qx^{2r_{1}/\sigma _{11}^{2}-2}(\sigma _{11}+\sigma _{12}x)^{-2r_{1}/\sigma _{11}^{2}-2} \exp \left\{ \frac{2r_{1}(\sigma _{11}+K_{1}\sigma _{12})}{K_{1}\sigma _{11}\sigma _{12}(\sigma _{11}+\sigma _{12}x)}\right\} , \end{aligned}$$

where Q is a constant such that $\int _{0}^{\infty }\varphi (x)\textrm{d}x=1$. From the ergodic theorem, it follows that

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{1}{t}\displaystyle \int \limits _{0}^{t}X(s)\textrm{d}s=\displaystyle \int \limits _{0}^{\infty }x\varphi (x)\textrm{d}x~~~a.s. \end{aligned}$$

Let X(t) be the solution of (C13) with $X(0)=x_{0}$, then, by the comparison theorem of one-dimensional stochastic differential equation (Peng and Zhu 2006), one has $x(t)\le X(t)$ for $t\ge 0$ a.s.

Applying the Itô’s formula yields

$$\begin{aligned} \begin{aligned} \textrm{d}(\ln y)&=\left[ r_{2}-\frac{r_{2}}{K_{2}}+\frac{e_{1}m_{1}x}{a_{1}+x}-\frac{m_{2}z}{a_{2}+y}-h_{1}-\frac{\sigma _{21}^{2}}{2}-\sigma _{21}\sigma _{22}y-\frac{\sigma _{22}^{2}}{2}y^{2}\right] \textrm{d}t\\&\quad \displaystyle +(\sigma _{21}+\sigma _{22}y)\text {d}B_{2}(t). \end{aligned}\nonumber \\ \end{aligned}$$

(C14)

Integrating both sides of (C14) from 0 to t and then dividing it by t leads to

$$\begin{aligned} \begin{aligned} \frac{\ln y(t)-\ln y_{0}}{t}&\le r_{2}-h_{1}-\frac{\sigma _{21}^{2}}{2}+\frac{e_{1}m_{1}}{t}\int _{0}^{t}\frac{x(s)}{a_{1}+x(s)}\textrm{d}s-\frac{\sigma _{22}^{2}}{2t}\int _{0}^{t}y^{2}(s)\textrm{d}s\\&\quad \displaystyle +\frac{\sigma _{21}B_{2}(t)}{t}+\frac{\sigma _{22}}{t}\int _{0}^{t}y(s)\textrm{d}B_{2}(s). \end{aligned} \end{aligned}$$

(C15)

Let $\tilde{M}(t)=\sigma _{22}\int _{0}^{t}y(s)\textrm{d}B_{2}(s)$, then $\tilde{M}(t)$ is a local martingale whose quadratic variations is $\langle \tilde{M},\tilde{M}\rangle _{t}=\sigma _{22}^{2}\int _{0}^{t}y^{2}(s)\textrm{d}s$. By employing the exponential martingales inequality (Mao 2007), for any positive constant $\delta$, one obtains

$$\begin{aligned} \mathbb {P}\left\{ \sup _{0\le t\le \delta }\left[ \tilde{M}(t)-\frac{1}{2}\langle \tilde{M},\tilde{M}\rangle _{t}\right] >2\ln \delta \right\} \le \frac{1}{\delta ^{2}}. \end{aligned}$$

From Borel-Cantelli Lemma (Mao 2007), it follows that, for almost all $\omega \in \Omega$, there exists a $\delta _{0}=\delta _{0}(\omega )$ such that

$$\begin{aligned} \sup _{0\le t\le \delta }\left[ \tilde{M}(t)-\frac{1}{2}\langle \tilde{M},\tilde{M}\rangle _{t}\right] \le 2\ln \delta . \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} \tilde{M}(t)\le 2\ln \delta +\frac{1}{2}\langle \tilde{M},\tilde{M}\rangle _{t}=2\ln \delta +\frac{\sigma _{22}^{2}}{2}\int _{0}^{t}y^{2}(s)\text {d}s,~~0\le t\le \delta ,~~\delta \ge \delta _{0},~~a.s. \end{aligned} \end{aligned}$$

Hence, for $0\le \delta -1\le t<\delta$, the following inequality holds:

$$\begin{aligned} \begin{aligned} \frac{\ln y(t)-\ln y_{0}}{t}&\le r_{2}-h_{1}-\frac{\sigma _{21}^{2}}{2}+\frac{e_{1}m_{1}}{t}\int _{0}^{t}\frac{X(s)}{a_{1}+X(s)}\textrm{d}s+\frac{\sigma _{21}B_{2}(t)}{t}+\frac{2\ln \delta }{\delta -1}.\\ \end{aligned}\nonumber \\ \end{aligned}$$

(C16)

Taking the superior limit on both sides of (C16) and using $\lim _{t\rightarrow \infty }{B_{2}(t)}/{t}=0$ a.s., by the strong law of large numbers for local martingale (Mao 2007), one gets

$$\begin{aligned} \limsup _{t\rightarrow \infty }\frac{\ln y(t)}{t}\le r_{2}-h_{1}-\frac{\sigma _{21}^{2}}{2}+e_{1}m_{1}\int _{0}^{\infty }\frac{x\varphi (x)}{a_{1}+x}\textrm{d}x<0, \end{aligned}$$

that is, $\lim _{t\rightarrow \infty }y(t)=0$, and then $\lim _{t\rightarrow \infty }z(t)=0$ a.s. As a result, for any $\epsilon >0$, there exists an $\Omega _{\epsilon }\subset \Omega$ such that

$$\begin{aligned} \mathbb {P}(\Omega _{\epsilon })>1-\epsilon ,~~\frac{m_{1}xy}{a_{1}+x}\le \frac{m_{1}\epsilon x}{a_{1}+x}. \end{aligned}$$

Note that

$$\begin{aligned} \text {d}x(t)\ge \left[ r_{1}x\left( 1-\frac{x}{K_{1}}\right) -\frac{m_{1}\epsilon x}{a_{1}+x}\right] \text {d}t+x(\sigma _{11}+\sigma _{12}x)\text {d}B_{1}(t) \end{aligned}$$

and

$$\begin{aligned} \text {d}x(t)\le r_{1}x\left( 1-\frac{x}{K_{1}}\right) \text {d}t+x(\sigma _{11}+\sigma _{12}x)\text {d}B_{1}(t), \end{aligned}$$

the distribution of the process x(t) converges to the measure with the density $\varphi$. The proof is complete. $\square$

1.4 Proof of Theorem 13

Proof

Applying the Itô’s formula to $\ln x$ produces

$$\begin{aligned} \textrm{d}(\ln x)=\left[ r_{1}-\frac{r_{1}}{K_{1}}x-\frac{m_{1}y}{a_{1}+x}-\frac{\sigma _{11}^{2}}{2}-\sigma _{11}\sigma _{12}x-\frac{\sigma _{12}^{2}}{2}x^{2}\right] \textrm{d}t +(\sigma _{11}+\sigma _{12}x)\text {d}B_{1}(t).\nonumber \\ \end{aligned}$$

(C17)

Integrating (C17) from 0 to t and dividing it by t on both sides leads to

$$\begin{aligned} \frac{\ln x(t)-\ln x_{0}}{t}\le r_{1}-\frac{\sigma _{11}^{2}}{2}-\frac{\sigma _{12}^{2}}{2t}\int _{0}^{t}x^{2}(s)\textrm{d}s +\frac{\sigma _{11}B_{1}(t)}{t}+\frac{\sigma _{12}}{t}\int _{0}^{t}x(s)\textrm{d}B_{1}(s).\nonumber \\ \end{aligned}$$

(C18)

Let $M(t)=\sigma _{12}\int _{0}^{t}x(s)\textrm{d}B_{1}(s)$, then M(t) is a local martingale whose quadratic variations is $\langle M,M\rangle _{t}=\sigma _{12}^{2}\int _{0}^{t}x^{2}(s)\textrm{d}s$. By employing the exponential martingales inequality (Mao 2007), for any positive constant $\delta$, one obtains the following inequality

$$\begin{aligned} \mathbb {P}\left\{ \sup _{0\le t\le \delta }\left[ M(t)-\frac{1}{2}\langle M,M\rangle _{t}\right] >2\ln \delta \right\} \le \frac{1}{\delta ^{2}}. \end{aligned}$$

From Borel-Cantelli Lemma (Mao 2007), it follows that, for almost all $\omega \in \Omega$, there exists $\delta _{0}=\delta _{0}(\omega )$ such that

$$\begin{aligned} \sup _{0\le t\le \delta }\left[ M(t)-\frac{1}{2}\langle M,M\rangle _{t}\right] \le 2\ln \delta . \end{aligned}$$

It is trivial to show that

$$\begin{aligned} M(t)\le 2\ln \delta +\frac{1}{2}\langle M,M\rangle _{t}=2\ln \delta +\frac{\sigma _{12}^{2}}{2}\int _{0}^{t}x^{2}(s)\textrm{d}s,~~0\le t\le \delta ,~~\delta \ge \delta _{0},~~a.s. \end{aligned}$$

Hence, for $0\le \delta -1\le t<\delta$, the following inequality holds

$$\begin{aligned} \begin{aligned} \frac{\ln x(t)-\ln x_{0}}{t}&\le r_{1}-\frac{\sigma _{11}^{2}}{2}+\frac{\sigma _{11}B_{1}(t)}{t}+\frac{2\ln \delta }{\delta -1}.\\ \end{aligned} \end{aligned}$$

(C19)

Taking the superior limit on the both sides of (C19) and noting that $\lim _{t\rightarrow \infty }{B_{1}(t)}/{t}=0$ a.s. the strong law of large numbers for local martingale (Mao 2007) implies that

$$\begin{aligned} \limsup _{t\rightarrow \infty }\frac{\ln x(t)}{t}\le r_{1}-\frac{\sigma _{11}^{2}}{2}<0, \end{aligned}$$

then $\lim _{t\rightarrow \infty }x(t)=0$ a.s. Therefore, there exists a set $\Omega _{\epsilon }\subset \Omega$ such that

$$\begin{aligned} \mathbb {P}(\Omega _{\epsilon })>1-\epsilon ,~~\frac{e_{1}m_{1}xy}{a_{1}+x}\le \frac{e_{1}m_{1}\epsilon y}{a_{1}+\epsilon }<\frac{e_{1}m_{1}\epsilon }{a_{1}}y. \end{aligned}$$

Applying the Itô’s formula to $\ln y$, integrating it from 0 to t, and dividing it by t on both sides, one has

$$\begin{aligned}{} & {} \frac{\ln y(t)-\ln y_{0}}{t}\le r_{2}\nonumber \\{} & {} \quad +\frac{e_{1}m_{1}\epsilon }{a_{1}}-h_{1}-\frac{\sigma _{21}^{2}}{2}-\frac{\sigma _{22}^{2}}{2t}\int _{0}^{t}y^{2}(s)\textrm{d}s +\frac{\sigma _{21}B_{2}(t)}{t}\nonumber \\{} & {} \quad +\frac{\sigma _{22}}{t}\int _{0}^{t}y(s)\textrm{d}B_{2}(s). \end{aligned}$$

(C20)

Taking the superior limit on both sides of (C20) and using the arbitrariness of $\epsilon$ yield

$$\begin{aligned} \limsup _{t\rightarrow \infty }\frac{\ln y(t)}{t}\le r_{2}-h_{1}-\frac{\sigma _{21}^{2}}{2}<0, \end{aligned}$$

which implies that $\lim _{t\rightarrow \infty }y(t)=0$ and $\lim _{t\rightarrow \infty }z(t)=0$ a.s. The proof is complete.

$\square$