Modelling the effect of resource subsidy on a two-species predator-prey system under the influence of environmental noises

Abstract

Our ecosystem is full of various types of food chain systems. Most of the species have more than one food source with various kinds of possible interactions. In this work, we have studied a predator-prey-subsidy model in a rapidly fluctuating environment. As per our consideration predator exploits both a prey population and an allochthonous resource which is provided as a subsidy to the system exogenously, with Holling Type II response functions. Furthermore, we have introduced Gaussian white noise to the main parameters subject to coupling of a prey-predator pair with its environment. We have examined the existence of unique global positive solutions for both the deterministic and stochastic systems. Boundedness, uniform continuity, global attractiveness in mean of solutions of the stochastic system are proved. Conditions for extinction of prey and predator population are derived and the importance of environmental noise as well as subsidy for extinction is discussed through mathematical results and numerical findings. We have derived conditions for persistence of the proposed stochastic system. The results of persistence and extinction have been justified by numerical simulations. Numerically we have found that presence of subsidy can be the cause of survival of predator even there is no prey available for consumption which fact is different from the studies of those systems where additional food is provided for predator. We have also observed that very high input of subsidy can be a cause for extinction of prey population.

Appendices

Appendix A

Proof of Theorem 3.1

Since RHS of system (2.1) is continuous and locally Lipschitz on $\mathbb {R}_{+}^{3}$, the solution (x(t), s(t), y(t)) of system (2.1) exists uniquely on $[0, \tau ]$, where $\tau \in (0,\infty )$. Now from (2.1) with the initial conditions, we have from first equation:

$$\begin{aligned} x(t)= & {} x_0 \exp \int _{0}^{t} \left( g - \theta x(r) - \frac{\alpha y(r)}{x(r)+s(r)+h} \right) dr > 0, \\&\forall t\ge 0. \end{aligned}$$

From second equation of system (2.1), we have

$$\begin{aligned} \begin{aligned}&\frac{ds}{dt} + \left( d_{1} + \frac{\beta y}{x+s+h} \right) s = q \\&\quad \implies s(t) = \left( s_{0} + q \int _{0}^{t}e^{\int _{0}^{r}\left( d_{1} + \frac{\beta y}{x+s+h} \right) dr} dr \right) e^{-\int _{0}^{t}\left( d_{1} + \frac{\beta y}{x+s+h} \right) dr} \\&\quad > 0,\ \ \forall t\ge 0. \end{aligned} \end{aligned}$$

From third equation of system (2.1), we have

$$\begin{aligned} y(t) = y_0 \exp \int _{0}^{t} \left( \frac{\epsilon \alpha x + \eta \beta s}{x+s+h} - \delta \right) dr > 0, \ \ \forall t\ge 0. \end{aligned}$$

Hence the theorem. $\square$

Proof of Theorem 3.2

Since coefficients of system (2.2) satisfy local Lipschitz condition, hence for any initial value $(x_0,s_0, y_0) \in \mathbb {R}_{+}^{3}$ there is a unique local solution $x(t),s(t), y(t)$ for $t\in [0, \tau _{e})$, where $\tau _e$ is the explosion time. To show this is a global positive solution we need to show that $\tau _{e} = \infty$. Let $r_0 \ge 0$ be sufficiently large so that both $x_0$, $s_0$ and $y_0$ lie in the interval $\displaystyle \left[ \frac{1}{r_0} , r_0 \right]$. We define stopping time $(\tau _r)$ for each integer $r \ge r_0$ such that

$$\begin{aligned} \tau _r= & {} \inf \left\{ t \in [0, \tau _e) : \min \left\{ x(t), s(t), y(t) \right\} \right. \\< & {} \left. \frac{1}{r} \ \text {or} \ \max \left\{ x(t), s(t), y(t) \right\} > r \right\} , \end{aligned}$$

with $\inf \phi = \infty$ ( $\phi$ denotes the empty set). It is easy to observe that $\tau _r$ increases as $t \rightarrow \infty$. Here we set $\displaystyle \tau _{\infty } = \lim _{r \rightarrow \infty } \tau _r$, whence $\tau _\infty \le \tau _e$ a.s. If it can be proved that $\tau _\infty = \infty ,$ then it is easy to conclude that $\tau _e = \infty$ and $(x(t),s(t), y(t)) \in \mathbb {R}^{3}_{+}$ for all $t \ge 0$ almost surely. So, to complete the proof all we need to prove is that $\tau _\infty = \infty$. It can be proved by contradiction. Let if possible the statement be false, then there is a pair of constants $T > 0$ and $\epsilon \in (0, 1)$ such that

$$\begin{aligned} P \left\{ \tau _\infty \le T \right\} > \epsilon . \end{aligned}$$

So, there exists an integer $r_1 \ge r_0$ such that

$$\begin{aligned} P \left\{ \tau _\infty \le T \right\} > \epsilon , \ \ \forall r\ge r_1 . \end{aligned}$$

(A.1)

Now we define a $\mathcal {C}^3$-function $F : \mathbb {R}^{3}_{+} \longrightarrow \mathbb {R_{+}}$ by

$$\begin{aligned} F(x,s,y)= & {} (x + 1 - \log (x)) + (s + 1 - \log (s)) \\&+ (y + 1 - \log (y)) \end{aligned}$$

Since $(z + 1 - \log (z)) \ge 0 , \ \forall z >0$, it follows that F(x, s, y) is non negative. Let us apply apply Itô formula to get (defining $\nu = \max \{ \epsilon \alpha , \eta \beta \}$):

$$\begin{aligned}&dF(x,s,y) = \left[ \left( 1-\frac{1}{x}\right) x \left( g - \theta x - \frac{\alpha y}{x+s+h} \right) \right. \\&\qquad + \left( 1-\frac{1}{s}\right) s \left( \frac{q}{s} - d_{1} - \frac{\beta y}{x+s+h} \right) \\&\qquad \left. + \left( 1-\frac{1}{y}\right) y \left( \frac{\epsilon \alpha x + \eta \beta s}{x+s+h} - \delta \right) \right] dt \\&\qquad + \left[ (x-1)\sigma _{1} dw_{1} - (s-1) \sigma _{2} dw_{2}\right. \\&\qquad \left. - (y-1) \sigma _{3} dw_{3} \right] \\&\quad = \left[ (x-1) \left( g - \theta x - \frac{\alpha y}{x+s+h} \right) \right. \\&\qquad + (s-1) \left( \frac{q}{s} - d_{1} - \frac{\beta y}{x+s+h} \right) \\&\qquad \left. + (y-1) \left( \frac{\epsilon \alpha x + \eta \beta s}{x+s+h} - \delta \right) \right] dt \\&\qquad + \left[ (x-1)\sigma _{1} dw_{1} - (s-1) \sigma _{2} dw_{2} \right. \\&\qquad \left. - (y-1) \sigma _{3} dw_{3} \right] \\&\le \left( gx - g + \theta x + q -d_{1}s + d_{1} + \nu y \right. \\&\qquad \left. - \delta y + \frac{\sigma _{1}^{2}}{2} + + \frac{\sigma _{2}^{2}}{2} + \frac{\sigma _{3}^{2}}{2} \right) dt \\&\qquad + \left[ (x-1)\sigma _{1} dw_{1} - (s-1) \sigma _{2} dw_{2} \right. \\&\qquad \left. -(y-1) \sigma _{3} dw_{3} \right] \\&\quad = (a_{1}x + a_{2}s + a_{3}y + a_{4})dt + \left[ (x-1)\sigma _{1} dw_{1} \right. \\&\qquad \left. - (s-1) \sigma _{2} dw_{2} - (y-1) \sigma _{3} dw_{3} \right] , \end{aligned}$$

where $a_{1} = g+\theta$, $a_{2}= -d_{1}$, $a_{3}= \nu - \delta$ and $a_{4} = - \left( g - q - d_{1} - \frac{\sigma _{1}^{2}}{2} - \frac{\sigma _{2}^{2}}{2} - \frac{\sigma _{3}^{2}}{2} \right)$

$$\begin{aligned} \begin{aligned}&\therefore dF(x,s,y) \le \left[ 2a_{1}(x+1-\log x ) + 2a_{2}(s+1-\log s ) \right. \\&\quad \left. + 2a_{3}(y+1-\log y ) + a_{4} \right] dt \\&\quad + \left[ (x-1)\sigma _{1} dw_{1} - (s-1) \sigma _{2} dw_{2} - (y-1) \sigma _{3} dw_{3} \right] \end{aligned} \end{aligned}$$

Let $a_{5}= \max \left\{ 2a_{1}, 2a_{2}, 2a_{3}, a_{4} \right\}$ and $v_{1} \bigwedge v_{2} := \min \{ v_{1}, v_{2} \}$.

$$\begin{aligned}&\therefore dF(x,s,y) \le a_{5} (1+F(x,s,y)) + \left[ (x-1)\sigma _{1} dw_{1} \right. \\&\quad \left. - (s-1) \sigma _{2} dw_{2} - (y-1) \sigma _{3} dw_{3} \right] \end{aligned}$$

Hence for $t_1 \le T$,

$$\begin{aligned} \begin{aligned}&\int _{0}^{\tau _{r}\bigwedge t_{1}} d(F(x(t),s(t),y(t))) \\&\quad <\int _{0}^{\tau _{r}\bigwedge t_{1}} a_{5} ( 1+ F(x(t),s(t),y(t))) dt \\&\qquad + \int _{0}^{\tau _{r}\bigwedge t_{1}} (x-1) \sigma _{1} dw_{1} - \int _{0}^{\tau _{r}\bigwedge t_{1}} (s-1) \sigma _{2} dw_{2}\\&\qquad -\int _{0}^{\tau _{r}\bigwedge t_{1}} (y-1) \sigma _{3} dw_{3} \end{aligned} \end{aligned}$$

Taking expectation on both sides, we get

$$\begin{aligned} \begin{aligned}&E \left( F \left( x (t),s (t),y (t) \right) \right) \big |_{t=\tau _{r}\bigwedge t_{1}} \\&\quad \le F(x_0,s_0,y_0) + E \int _{0}^{\tau _{r}\bigwedge t_{1}} \\&\qquad a_{5} ( 1+ F(x(t),s(t),y(t))) dt\\&\quad \le F(x_0,s_0, y_0) + a_{5} t_1 \\&\qquad + a_{5} E \int _{0}^{\tau _{r}\bigwedge t_{1}} F(x(t),s(t),y(t)) dt \\&\quad \le F(x_0,s_0, y_0) + a_{5} T \\&\qquad + a_{5} \int _{0}^{t_1} E \left[ F \left( x \left( \tau _{r}\bigwedge t_{1} \right) ,\right. \right. \\&\quad \left. \left. s \left( \tau _{r}\bigwedge t_{1} \right) ,y \left( \tau _{r}\bigwedge t_{1} \right) \right) \right] dt \end{aligned} \end{aligned}$$

By Gronwall inequality [27]:

$$\begin{aligned} E(F( x(\tau _{r}\bigwedge t_{1}), s(\tau _{r}\bigwedge t_{1}),y(\tau _{r}\bigwedge t_{1}))) \le a_{6}, \end{aligned}$$

(A.2)

where $\displaystyle a_{6} = \left( F(x_0,s_0, y_0) + a_{5} T \right) e ^{a_{5}T}.$

Define $\displaystyle \Omega _{r} = \left\{ \tau _r \le T \right\}$ for $\displaystyle r \ge r_{1}$ and by (A.1), $P(\Omega _{r}) \ge \epsilon .$ Note that for every $\tau ^{\prime } \in \Omega _{r}$ there is at least one of $x(\tau _r,\tau ^{\prime }),s(\tau _r,\tau ^{\prime }), y(\tau _r, \tau ^{\prime } )$ which is equal either r or $\displaystyle \frac{1}{r}$. So $F\left( x\left( \tau _r, \tau ^{\prime }\right) ,s(\tau _r,\tau ^{\prime }), y\left( \tau _r, \tau ^{\prime }\right) \right)$ is not less than the smallest of

$$\begin{aligned} \displaystyle r + 1 - \log (r) \ \text {and} \ \frac{1}{r} +1 - \log \left( \frac{1}{r}\right) = \frac{1}{r} + 1 + \log (r). \end{aligned}$$

Consequently,

$$\begin{aligned}&F(x(\tau _r, \tau ^{\prime }),s(\tau _r,\tau ^{\prime }), y(\tau _r, \tau ^{\prime })) \\&\quad \ge \left( r + 1 - \log (r) \right) \bigwedge \left( \frac{1}{r} + 1 + \log (r) \right) . \end{aligned}$$

Now from (A.1) and (A.2), we get

$$\begin{aligned} \begin{aligned} a_6&\ge E [ 1_{\Omega _{r}} F (x(\tau _{r} , \tau ^{\prime }),s(\tau _{r} , \tau ^{\prime }), y(\tau _{r} , \tau ^{\prime }))] \\&\ge \epsilon \left[ \left( r + 1 - \log (r) \right) \bigwedge \left( \frac{1}{r} + 1 + \log (r) \right) \right] , \end{aligned} \end{aligned}$$

where $1_{\Omega _{r}}$ is the indicator function of $\Omega _{r}$. Thus $r \rightarrow \infty$ leads towards the contradiction $\infty > a_{6} = \infty$. Hence our assumption was wrong. So, $\tau _{\infty } = \infty$. $\square$

Appendix B

Proof of Theorem 4.1

From (2.2):

$$\begin{aligned} dx = x \left( g - \theta x - \frac{\alpha y}{x+s+h} \right) dt + \sigma _{1} x dw_{1} \end{aligned}$$

Let $V_{1}(t) = e^{t}x^{p}$ and apply Itô formula:

$$\begin{aligned} \begin{aligned} dV_{1}(t)&= \left[ e^{t}x^{p} + p e^{t}x^{p-1} x \left( g - \theta x - \frac{\alpha y}{x+s+h} \right) \right. \\&\quad \left. + \frac{p(p-1)}{2}e^{t}x^{p} \sigma _{1}^{2} \right] dt + pe^{t}x^{p} \sigma _{1} dw_{1} \\&= pe^{t}x^{p} \left[ \frac{1}{p} + g - \theta x - \frac{\alpha y}{x+s+h} + \frac{p-1}{2}\sigma _{1}^{2} \right] dt \\&\quad + pe^{t}x^{p} \sigma _{1} dw_{1} \end{aligned} \end{aligned}$$

Taking expectation:

$$\begin{aligned}&E(V_{1}(t)) \le x_{0}^{p} + p \int _{0}^{t} e^{r} E \left[ x^{p} \left( \frac{1}{p} + g - \theta x \right. \right. \\&\quad \left. \left. + \frac{p-1}{2}\sigma _{1}^{2} \right) \right] dr . \end{aligned}$$

Let us consider

$$\begin{aligned} f(x) = x^{p} \left( \frac{1}{p} + g - \theta x + \frac{p-1}{2}\sigma _{1}^{2} \right) \end{aligned}$$

Now,

$$\begin{aligned} f^{\prime }(x) = 0 \implies x = \frac{p \left( \frac{1}{p} + g + \frac{p-1}{2}\sigma _{1}^{2} \right) }{\theta (p+1)} = x^{*} \ (\text {say}) . \end{aligned}$$

Calculating at that point we found $f^{\prime \prime }(x^{*}) < 0$. So, f(x) has maximum at $\displaystyle x = x^{*}.$

Hence,

$$\begin{aligned} \begin{aligned} f\big |_{\max }&= \left[ \frac{p\left( \frac{1}{p}+g+\frac{p-1}{2}\sigma _{1}^{2} \right) }{\theta (p+1)} \right] ^{p}\\&\quad \left( \frac{1}{p+1} \left( \frac{1}{p}+g+\frac{p-1}{2}\sigma _{1}^{2} \right) \right) \\&= \left( \frac{p}{\theta }\right) ^{p} \left[ \frac{\frac{1}{p}+g+\frac{p-1}{2}\sigma _{1}^{2}}{p+1} \right] ^{p+1} \end{aligned} \end{aligned}$$

i.e.,

$$\begin{aligned} \begin{aligned}&E(V_{1}(t)) \le x_{0}^{p} + p \left( \frac{p}{\theta }\right) ^{p} \left( \frac{\frac{1}{p}+g+\frac{p-1}{2}\sigma _{1}^{2}}{p+1} \right) ^{p+1} (e^{t}-1) \\&\quad \implies E(x^{p}) \le \left[ x_{0}^{p} - p \left( \frac{p}{\theta }\right) ^{p} \left( \frac{\frac{1}{p}+g+\frac{p-1}{2}\sigma _{1}^{2}}{p+1} \right) ^{p+1} \right] e^{-t} \\&\quad + p \left( \frac{p}{\theta }\right) ^{p} \left( \frac{\frac{1}{p}+g+\frac{p-1}{2}\sigma _{1}^{2}}{p+1} \right) ^{p+1} \end{aligned} \end{aligned}$$

For $t=0$, $E(x^{p}) \le x_{0}^{p}$ and for $t\rightarrow \infty$, $\displaystyle E(x^{p}) \le p \left( \frac{p}{\theta }\right) ^{p} \left( \frac{\frac{1}{p}+g+\frac{p-1}{2}\sigma _{1}^{2}}{p+1} \right) ^{p+1}.$

$$\begin{aligned}&\text {Let} \ U_{1}(p)= \max \left\{ x_{0}^{p}, p\left( \frac{p}{\theta }\right) ^{p} \left[ \frac{\frac{1}{p}+g + \frac{p-1}{2}\sigma _{1}^{2}}{p+1} \right] ^{p+1} \right\} . \nonumber \\&\quad \text {So}, \ E(x^{p}) \le U_{1}(p). \end{aligned}$$

(B.1)

Now from second equation of system (2.2), we have

$$\begin{aligned} \begin{aligned} ds&= \left( q - d_{1}s - \frac{\beta sy}{x+s+h} \right) dt - \sigma _{2} s dw_{2} \\&\le \left( q - d_{1}s \right) dt - \sigma _{2} s dw_{2} \end{aligned} \end{aligned}$$

Let $\phi (t)$ be the unique positive solution satisfying the following equation:

$$\begin{aligned} d\phi (t) = \left( q - d_{1}\phi (t) \right) dt - \sigma _{2} \phi (t) dw_{2} \ \text {with} \ \phi (0)= s_{0} . \end{aligned}$$

We consider, $V_{2}(t) = e^{t}\phi ^{p}$ and apply Itô formula:

$$\begin{aligned} \begin{aligned} d(e^{t}\phi ^{p})&= \left[ e^{t}\phi ^{p} + p e^{t}\phi ^{p-1} (q - d_{1}\phi )+ \frac{p(p-1)}{2} e^{t}\phi ^{p} \sigma _{2}^{2} \right) dt\\&\quad - p e^{t}\phi ^{p} \sigma _{2} dw_{2} \\&= p e^{t}\phi ^{p-1} \left[ q + \phi \left( \frac{1}{p} - d_{1} + \frac{p-1}{2}\sigma _{2}^{2} \right) \right] dt \\&\quad - p e^{t}\phi ^{p}\sigma _{2} dw_{2} \end{aligned} \end{aligned}$$

Taking expectation on both sides, we get

$$\begin{aligned}&E(e^{t}\phi ^{p}) \le \phi (0)^{p} + p \int _{0}^{t} e^{r} E\left[ \phi ^{p-1}\left\{ q+ \phi \left( \frac{1}{p}\right. \right. \right. \\&\quad \left. \left. \left. -d_{1} + \frac{p-1}{2}\sigma _{2}^{2} \right) \right\} \right] dr. \end{aligned}$$

Let $\displaystyle g(\phi ) = \phi ^{p-1}\left\{ q+ \phi \left( \frac{1}{p}-d_{1} + \frac{p-1}{2}\sigma _{2}^{2} \right) \right\}$ $\displaystyle \therefore g^{\prime }(\phi ) = 0 \ \implies \ \phi (t) = \frac{q(1-p)}{p \left( \frac{1}{p}-d_{1} + \frac{p-1}{2}\sigma _{2}^{2} \right) }.$

Hence, $\displaystyle g\big |_{\max } = \left( \frac{q}{p}\right) ^{p} \left[ \frac{1-p}{\frac{1}{p}-d_{1} + \frac{p-1}{2}\sigma _{2}^{2}} \right] ^{p-1}.$ Same as calculated previously we can show that $\displaystyle E(\phi ) \le \max \left\{ \phi (0)^{p}, p \left( \frac{q}{p}\right) ^{p} \left[ \frac{1-p}{\frac{1}{p}-d_{1} + \frac{p-1}{2}\sigma _{2}^{2}} \right] ^{p-1} \right\}$.

Since $s(t) \le \phi (t)$ and $s_{0}=\phi (0)$, so by comparison theorem, we can conclude that

$$\begin{aligned}&E(s(t)) \le U_{2}(t), \ \text {where} \ U_{2}(t) \nonumber \\&\quad = \max \left\{ s_{0}^{p}, p \left( \frac{q}{p}\right) ^{p} \left[ \frac{1-p}{\frac{1}{p}-d_{1} + \frac{p-1}{2}\sigma _{2}^{2}} \right] ^{p-1} \right\} . \end{aligned}$$

(B.2)

Again, from third equation of system (2.2), we have

$$\begin{aligned} dy = y \left( \frac{\epsilon \alpha x + \eta \beta s}{x+s+h} - \delta \right) dt - \sigma _{3} y dw_{3}. \end{aligned}$$

Let us consider $\displaystyle V_{3}(t) = \log _{e} y(t)$ and apply Itô formula:

$$\begin{aligned} \begin{aligned} d(\log y(t))&= \left( \frac{\epsilon \alpha x + \eta \beta s}{x+s+h} - \delta - \frac{\sigma _{3}^{2}}{2} \right) dt - \sigma _{3} dw_{3} \\&\le \left( \nu - \delta - \frac{\sigma _{3}^{2}}{2} \right) dt - \sigma _{3} dw_{3}, \\&\quad \text {where} \ \nu = \max \left\{ \epsilon \alpha , \eta \beta \right\} \\ \implies \log y(t)&\le \log y_{0} + \left( \nu - \delta - \frac{\sigma _{3}^{2}}{2} \right) t - M_{3} \\ \implies y(t)&\le y_{0} e^{\left( \nu - \delta - \frac{\sigma _{3}^{2}}{2} \right) t - M_{3}} \\ \implies y^{p}(t)&\le y_{0}^{p} e^{p \left( \nu - \delta - \frac{\sigma _{3}^{2}}{2} \right) t - p M_{3}} \end{aligned} \end{aligned}$$

Taking expectation (by Lemma 4.1):

$$\begin{aligned} E(y^{p}(t))&\le y_{0}^{p} E\left( e^{p \left( \nu - \delta - \frac{\sigma _{3}^{2}}{2} \right) t - p M_{3}} \right) \nonumber \\&= y_{0}^{p} e^{p \left( \nu - \delta - \frac{\sigma _{3}^{2}}{2} \right) t + \frac{p^{2}\sigma _{3}^{2}}{2}t} \nonumber \\&= y_{0}^{p} e^{p \left( \nu - \delta + \frac{p-1}{2}\sigma _{3}^{2} \right) t } \nonumber \\&\text {Now for} \ \nu + \frac{p-1}{2}\sigma _{3}^{2} < \delta , \ E(y^{p}(t) \le y_{0}^{p} \end{aligned}$$

(B.3)

Hence, (B.1) and (B.3) conclude the theorem. $\square$

Appendix C

Proof of Theorem 5.1

From (2.2), we have

$$\begin{aligned} dx = x \left( g - \theta x - \frac{\alpha y}{x+s+h} \right) dt + \sigma _{1} x dw_{1}. \end{aligned}$$

Let us take $F_{1}(t) = \log x(t)$ and apply Itô formula

$$\begin{aligned} \begin{aligned} d(\log x(t))&= \left( g - \theta x - \frac{\alpha y }{x+s+h} - \frac{\sigma _{1}^{2}}{2} \right) dt + \sigma _{1} dw_{1} \\&\le \left( g - \frac{\sigma _{1}^{2}}{2} \right) dt + \sigma _{1} dw_{1} \end{aligned} \end{aligned}$$

Integrating and dividing both sides by t, we have

$$\begin{aligned} \begin{aligned}&\frac{\log x(t)}{t} \le \frac{\log x_{0}}{t} + g - \frac{\sigma _{1}^{2}}{2} + \frac{M_{1}}{t} \\&\quad \implies \limsup _{t\rightarrow \infty } \frac{\log x(t)}{t} = g - \frac{\sigma _{1}^{2}}{2} \end{aligned} \end{aligned}$$

Hence, $\displaystyle \lim _{t\rightarrow \infty } x(t) = 0 \ \ a.s. \ if \ \ g< \frac{\sigma _{1}^{2}}{2}$.

Therefore, for every $\epsilon ^{\prime }> 0,\ \exists \ t_{0} ( > 0 )$ and $\Omega _{\epsilon ^{\prime }}$ s.t. $P(\Omega _{\epsilon ^{\prime }}) \ge 1- \epsilon ^{\prime }$ and $x < \epsilon ^{\prime },\ \forall t \ge t_{0},\ x \in \Omega _{\epsilon ^{\prime }}.$

Now, from last equation of system (2.2), we have

$$\begin{aligned} \begin{aligned} dy = y \left( \frac{\epsilon \alpha x + \eta \beta s}{x+s+h} - \delta \right) dt - \sigma _{3} y dw_{3} \end{aligned} \end{aligned}$$

Proceeding as before:

$$\begin{aligned} \begin{aligned} d(\log y(t))&= \left( \frac{\epsilon \alpha x + \eta \beta s}{x+s+h} - \delta - \frac{\sigma _{3}^{2}}{2} \right) dt - \sigma _{3} dw_{3} \\&\le \left( \frac{\epsilon \alpha x + \eta \beta s}{h} - \delta - \frac{\sigma _{3}^{2}}{2} \right) dt - \sigma _{3} dw_{3} \\&\le \left( \frac{\epsilon \alpha \epsilon ^{\prime } + \eta \beta s_{0}}{h} - \delta - \frac{\sigma _{3}^{2}}{2} \right) dt \\&\quad -\sigma _{3} dw_{3} \ [\because x < \epsilon ^{\prime } \ \text {and using} \ Theorem~4.1.] \\&\quad \therefore \frac{\log y(t)}{t} \le \frac{\log y_{0}}{t} \\&\quad + \left( \frac{\epsilon \alpha \epsilon ^{\prime } + \eta \beta s_{0}}{h} - \delta - \frac{\sigma _{3}^{2}}{2} \right) - \frac{M_{3}}{t} \\&\implies \limsup _{t\rightarrow \infty } \frac{\log y(t)}{t} \le \frac{\epsilon \alpha \epsilon ^{\prime } + \eta \beta s_{0}}{h} - \delta - \frac{\sigma _{3}^{2}}{2} \end{aligned} \end{aligned}$$

Since $\epsilon ^{\prime }$ is arbitrarily small, so we can conclude that

$$\begin{aligned} \displaystyle \lim _{t\rightarrow \infty } y(t) = 0 \ \ a.s. \ \text {if} \ \ \frac{\eta \beta s_{0}}{h}< \delta + \frac{\sigma _{3}^{2}}{2}. \end{aligned}$$

Hence the theorem. $\square$

Appendix D

Proof of Theorem 6.1

From (2.2):

$$\begin{aligned} dx = x \left( g - \theta x - \frac{\alpha y}{x+s+h} \right) dt + \sigma _{1} x dw_{1} \end{aligned}$$

We take $\displaystyle \Gamma _{1}(x(t),y(t)) := x \left( g - \theta x - \frac{\alpha y}{x+s+h} \right)$ and $\displaystyle \Gamma _{2}(x(t),y(t)) := \sigma _{1} x.$

Hence, $\displaystyle dx = \Gamma _{1}(x(t),s(t),y(t)) dt + \Gamma _{2}(x(t),s(t),y(t)) dw_{1}.$

Using Theorem 4.1:

$$\begin{aligned} \begin{aligned}&E\left| \Gamma _{1}(x(t),s(t),y(t))\right| ^{p} \\&\quad \le \frac{1}{2} E |x|^{2p} + \frac{1}{2} E \left| (g + \theta x + \alpha y \right| ^{2p} \ [\because A.M. \ge G.M.] \\&\quad \le \frac{1}{2} E (x^{2p}) + \frac{4^{2p-1}}{2} \left[ g^{2p} + \theta ^{2p} E(x^{2p}) + \alpha ^{2p} E(y^{2p}) \right] \\&\quad = \left( \frac{1}{2} + \frac{4^{2p-1}}{2} \theta ^{2p} \right) E(x^{2p}) \\&\qquad + \frac{4^{2p-1}}{2} g^{2p} + \frac{4^{2p-1}}{2} \alpha ^{2p} E(y^{2p}) \\&\quad \le \left( \frac{1}{2} + \frac{4^{2p-1}}{2} \theta ^{2p} \right) U_{1}(2p) + \frac{4^{2p-1}}{2} g^{2p}\\&\qquad + \frac{4^{2p-1}}{2}\alpha ^{2p} y_{0}^{2p} \\&\quad = K_{1}(p) \ (\mathrm {say}) \end{aligned} \end{aligned}$$

Also, $\displaystyle E\left| \Gamma _{2}(x(t),s(t),y(t))\right| ^{p} \le \sigma _{1}^{p} U_{1}(p) = K_{2}(p) \ (\mathrm {say})$

Now, we write the differential equation in its scholastic integral form:

$$\begin{aligned} x(t)= & {} x_{0} + \int _{0}^{t} \Gamma _{1}(x(r),s(r),y(r)) dr\\&+\int _{0}^{t} \Gamma _{2}(x(r),s(r),y(r)) dw_{1}(r) \end{aligned}$$

From moment inequality [27] of Itô for $0 \le t_1< t_2 < \infty$ and $p \ge 2$, we have

$$\begin{aligned}&E \left| \int _{t_{1}}^{t_{2}} \Gamma _{2}(x(r),s(r),y(r)) dw_{1}(r) \right| ^{p} \nonumber \\&\quad \le \left[ \frac{p(p-1)}{2} \right] ^{\frac{p}{2}} (t_2 - t_1)^{\frac{p-2}{2}} \int _{t_{1}}^{t_{2}} \left| \Gamma _{2}(x(r),\right. \nonumber \\&\qquad \left. s(r),y(r)) \right| ^{p} dr \end{aligned}$$

(D.1)

Now, we apply Hölder’s inequality for $t_{2} - t_{1} \le 1$ and using (D.1),

$$\begin{aligned} \begin{aligned}&E|x(t_2)-x(t_{1})|^{p} \le 2^{p-1}E \left[ \int _{t_1}^{t_2} \Gamma _{1}(x(r),s(r),y(r)) dr \right] ^{p} \\&\qquad + 2^{p-1}E \left[ \int _{t_1}^{t_2} \Gamma _{2}(x(r),s(r),y(r)) dw_{1}(r) \right] ^{p} \\&\le 2^{p-1} (t_{2}-t_{1})^{(p-1)} \int _{t_1}^{t_2} K_{1}(p) ds \\&\qquad +2^{p-1}\left[ \frac{p(p-1)}{2} \right] ^{\frac{p}{2}} (t_2 - t_1)^{\frac{p-2}{2}} \int _{t_{1}}^{t_{2}} K_{2}(p) ds\\&= 2^{p-1} (t_{2}-t_{1})^{p} K_{1}(p) + 2^{p-1}\left[ \frac{p(p-1)}{2} \right] ^{\frac{p}{2}} (t_{2}-t_{1})^{\frac{p}{2}} K_{2}(p) \\&\le 2^{p-1} (t_{2}-t_{1})^{\frac{p}{2}} \left[ 1+ \frac{p(p-1)}{2} \right] \left( K_{1}(p) + K_{2}(p) \right) \end{aligned} \end{aligned}$$

Hence, it can be concluded that every sample path of x(t) is locally but uniformly Hölder continuous with exponent $\gamma \in \left( 0 , \frac{p-2}{2p} \right)$, i.e, every sample path of x(t) is uniformly continuous on $t\ge 0$. Similarly, it can be shown that every sample path of s(t) and y(t) is uniformly continuous on $t\ge 0$.

Proof of Theorem 6.2

We know that

$$\begin{aligned} d(\log x(t)) = \left( g - \theta x - \frac{\alpha y }{x+s+h} - \frac{\sigma _{1}^{2}}{2} \right) dt + \sigma _{1} dw_{1} \end{aligned}$$

Now, integrating both side and dividing by t, we get

$$\begin{aligned}&\frac{\log x(t) - \log x_{0}}{t} = \left( g - \frac{\sigma _{1}^{2}}{2} \right) - \frac{\theta }{t}\int _{0}^{t} x(r)dr \\&\quad - \frac{\alpha }{t} \int _{0}^{t}\frac{y(r)}{x(r) +s(r) +h} dr + \frac{M_{1}}{t} = H_{1}(t)(\text {say}). \end{aligned}$$

Applying Itô formula on the one dimensional stochastic system (6.1), we get

$$\begin{aligned} d(\log z(t)) = \left( g - \theta z - \frac{\sigma _{1}^{2}}{2} \right) dt + \sigma _{1} dw_{1} \end{aligned}$$

Now, integrating both side and dividing by t, we get

$$\begin{aligned} \frac{\log z(t) - \log z_{0}}{t}= & {} \left( g - \frac{\sigma _{1}^{2}}{2} \right) - \frac{\theta }{t}\int _{0}^{t} z(r)dr \\&+ \frac{M_{1}}{t} = H_{2}(t)(\text {say}). \end{aligned}$$

Clearly, $H_{1}(t) \le H_{2}(t)$

$$\begin{aligned} \therefore \frac{\theta }{t} \int _{0}^{t} (z(r)-x(r))dr\le & {} \frac{\alpha }{t} \int _{0}^{t} \frac{y(r)}{x(r) +s(r) +h} dr \nonumber \\\le & {} \frac{\alpha }{th} \int _{0}^{t} y(r) dr \end{aligned}$$

(D.2)

Applying Itô formula on third equation of system (2.2), we have

$$\begin{aligned} d(\log y(t))&= \left[ \frac{\epsilon \alpha x + \eta \beta s}{x+s+h} - \delta - \frac{\sigma _{3}^{2}}{2} \right] dt - \sigma _{3} dw_{3} \\&\ge \left[ \frac{\epsilon \alpha x }{x+s_{0}+h} - \delta - \frac{\sigma _{3}^{2}}{2} \right] dt - \sigma _{3} dw_{3} \\&= \left[ \frac{\epsilon \alpha z }{z+s_{0}+h} - \left( \frac{\epsilon \alpha z }{z+s_{0}+h} - \frac{\epsilon \alpha x }{x+s_{0}+h} \right) \right. \\&\quad \left. - \delta - \frac{\sigma _{3}^{2}}{2} \right] dt - \sigma _{3} dw_{3} \\&= \left[ \frac{\epsilon \alpha z }{z+s_{0}+h} - \left( \frac{\epsilon \alpha (s_0+h) (z-x) }{(z+s_{0}+h) (x+s_{0}+h)} \right) \right. \\&\quad \left. - \delta - \frac{\sigma _{3}^{2}}{2} \right] dt - \sigma _{3} dw_{3} \\&\ge \left[ \frac{\epsilon \alpha z }{z+s_{0}+h} - \frac{\epsilon \alpha }{(s_0+h)}(z-x) - \delta - \frac{\sigma _{3}^{2}}{2} \right] dt\\&\quad - \sigma _{3} dw_{3} \end{aligned}$$

Integrating and dividing both sides by t, we get

$$\begin{aligned} \begin{aligned}&\frac{\log y(t) - \log y_{0}}{t} \ge \frac{\epsilon \alpha }{t} \int _{0}^{t} \frac{z(r)}{z(r)+s_{0}+h}dr \\&\qquad - \frac{\epsilon \alpha }{(s_0+h)t} \int _{0}^{t}(z(r)-x(r))dr - \delta - \frac{\sigma _{3}^{2}}{2} - \frac{M_{3}}{t} \\&\quad \ge \frac{\epsilon \alpha }{t} \int _{0}^{t} \frac{z(r)}{z(r)+s_{0}+h} dr \\&\qquad - \frac{\epsilon \alpha ^{2}}{(s_0+h)h \theta t} \int _{0}^{t}y(r)dr - \delta - \frac{\sigma _{3}^{2}}{2} - \frac{M_{3}}{t} \\&= \epsilon \alpha \int _{0}^{\infty } \frac{z}{z+s_{0}+h} \mu (x) dx - \epsilon ^{\prime } - \frac{\epsilon \alpha ^{2}}{(s_0+h)h \theta } \left\langle y \right\rangle _{t}\\&\qquad - \delta - \frac{\sigma _{3}^{2}}{2}- \frac{M_{3}}{t} \\&= \Delta - \epsilon ^{\prime } - \frac{\epsilon \alpha ^{2}}{(s_0+h)h \theta } \left\langle y \right\rangle _{t} - \frac{M_{3}}{t}. \end{aligned} \end{aligned}$$

For sufficiently large t, applying Lemma 6.2 (b) and using arbitrariness of $\epsilon ^{\prime }$ it can be derived that

$$\begin{aligned} \liminf _{t\rightarrow \infty } \left\langle y \right\rangle _{t} = \frac{\Delta (s_0+h)h \theta }{\epsilon \alpha ^{2}}, \ \text {a.s.} \end{aligned}$$

So, system (2.2) will be persistent in mean if $\Delta > 0$. Hence the theorem. $\square$

Appendix E

Proof of Theorem 7.1

Let us consider $F(t) = |x_{1}(t)-x_{2}(t)| + \theta _{1}|s_{1}(t)-s_{2}(t)| + \theta _{2} |y_{1}(t)-y_{2}(t)|$ and apply Itô formula on system (2.2), where $\theta _{1}$ and $\theta _{2}$ are real constants which can be chosen as requirement.

$$\begin{aligned} \begin{aligned}&L(F(t)) = sgn \left( x_{1}(t) - x_{2}(t) \right) \left[ gx_{1} - \theta x_{1}^{2} - \frac{\alpha x_{1}y_{1}}{x_{1}+s_{1}+h} \right. \\&\qquad \left. - gx_{2} + \theta x_{2}^{2} + \frac{\alpha x_{2}y_{2}}{x_{2}+s_{2}+h} \right] dt \\&\quad + \theta _{1} sgn \left( s_{1}(t) - s_{2}(t) \right) \left[ -d_{1}s_{1} - \frac{\beta s_{1}y_{1}}{x_{1}+s_{1}+h} \right. \\&\qquad \left. + d_{1}s_{2} + \frac{\beta s_{2}y_{2}}{x_{2}+s_{2}+h} \right] dt \\&\qquad + \theta _{2} sgn \left( y_{1}(t) - y_{2}(t) \right) \left[ \frac{\epsilon \alpha x_{1}y_{1} + \eta \beta s_{1}y_{1}}{x_{1}+s_{1}+h} - \delta y_{1}\right. \\&\qquad \left. - \frac{\epsilon \alpha x_{2}y_{2} + \eta \beta s_{2}y_{2}}{x_{2}+s_{2}+h} + \delta y_{2} \right] dt \\&\quad = sgn \left( x_{1}(t) - x_{2}(t) \right) \left[ g(x_{1} - x_{2})- \theta (x_{1}+x_{2})(x_{1}-x_{2})\right. \\&\qquad \left. - \alpha \frac{x_{1}y_{1} (x_{2}+s_{2}+h) - x_{2}y_{2}(x_{1}+s_{1}+h)}{(x_{1}+s_{1}+h)(x_{2}+s_{2}+h)} \right] dt\\&\qquad + \theta _{1} sgn \left( s_{1}(t) - s_{2}(t) \right) \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned}&\left[ -d_{1}(s_{1} - s_{2}) - \beta \frac{s_{1}y_{1}(x_{2}+s_{2}+h) - s_{2}y_{2}(x_{1}+s_{1}+h)}{(x_{1}+s_{1}+h)(x_{2}+s_{2}+h)} \right] dt + \theta _{2} sgn \left( y_{1}(t) \right. \\&\qquad \left. - y_{2}(t) \right) \left[ \frac{(\epsilon \alpha x_{1}y_{1} + \eta \beta s_{1}y_{1})(x_{2}+s_{2}+h) - (\epsilon \alpha x_{2}y_{2} + \eta \beta s_{2}y_{2})(x_{1}+s_{1}+h) }{(x_{1}+s_{1}+h)(x_{2}+s_{2}+h)} \right. \\&\quad \left. - \delta (y_{1}- y_{2}) \right] dt \\&\quad = \left\{ \left[ g - \theta (x_{1}+x_{2}) - \frac{\alpha (y_2s_2+hy_1)}{(x_{1}+s_{1}+h)(x_{2}+s_{2}+h)} - \frac{\theta _1 \beta s_2 y_2}{(x_{1}+s_{1}+h)(x_{2}+s_{2}+h)} \right. \right. \\&\qquad \left. \left. + \theta _{2} \frac{\epsilon \alpha (y_2s_2+y_1h)+\eta \beta s_{2}y_{2}}{(x_{1}+s_{1}+h)(x_{2}+s_{2}+h)} \right] |x_1-x_2|+ \left[ - d_1 \theta _1 \right. \right. \\&\qquad \left. \left. -\frac{\alpha x_2 y_2 + \theta _1 \beta (x_2 y_2 + y_1 h) - \theta _2 \epsilon \alpha x_2 y_2 - \theta _2 \eta \beta (x_2 y_2 + y_1 h)}{(x_{1}+s_{1}+h)(x_{2}+s_{2}+h)} \right] |s_1-s_2| + \left[ - \delta \theta _2 \right. \right. \\&\qquad \left. \left. -\frac{\alpha (x_1x_2+x_1s_2+x_2h)+\theta _1 \beta (s_1s_2+s_1x_2+s_2h) - \theta _2 \alpha \epsilon (x_1x_2+x_1s_2+x_2h)}{(x_{1}+s_{1}+h)(x_{2}+s_{2}+h)} \right. \right. \\&\qquad \left. \left. - \frac{\theta _2 \eta \beta (s_1s_2+s_1x_2+s_2h)}{(x_{1}+s_{1}+h)(x_{2}+s_{2}+h)} \right] |y_1-y_2| \right\} dt \\&\quad \le \left\{ \left[ g- 2\theta m - \frac{(y_2s_2+hy_1)(\alpha - \theta _2 \epsilon \alpha ) + s_2y_2(\theta _1 \beta - \eta \beta \theta _2)}{(x_{1}+s_{1}+h)(x_{2}+s_{2}+h)} \right] |x_1-x_2| \right. \\&\qquad \left. + \left[ - d_1 \theta _1 - \frac{(\alpha -\theta _2\epsilon \alpha )x_2y_2 + (\theta _1\beta -\theta _2 \eta \beta )(x_2y_2+y_1h)}{(x_{1}+s_{1}+h)(x_{2}+s_{2}+h)} \right] |s_1-s_2| \right. \\&\qquad \left. + \left[ - \delta \theta _2 - \frac{N}{(x_{1}+s_{1}+h)(x_{2}+s_{2}+h)} \right] |y_1-y_2| \right\} dt \\&\qquad [\because m\le E(x(t)), m \ \textit{is as defined in}~4.2 ] \end{aligned} \end{aligned}$$

where $N= (\alpha -\theta _2\epsilon \alpha )(x_1x_2+x_1s_2+x_2h) + (\theta _1\beta -\theta _2 \eta \beta )(s_1s_2+s_1x_2+s_2h)$.

Now we chose $\displaystyle \theta _{1} = \frac{\eta }{\epsilon }$ and $\displaystyle \theta _{2} = \frac{1}{\epsilon }$

$$\begin{aligned} \begin{aligned}&\therefore L(F(t)) \le \left[ (g-2\theta m)|x_1-x_2| - \frac{d_1\eta }{\epsilon }|s_1-s_2| \right. \\&\quad \left. - \frac{\delta }{\epsilon }|y_1-y_2| \right] dt \\&\quad \implies F(t) \le F(0) + (g-2\theta m) \int _{0}^{t} |x_{1}(r) - x_{2}(r)|dr \\&\quad - \frac{d_1\eta }{\epsilon } \int _{0}^{t} |s_{1}(r) - s_{2}(r)|dr - \frac{\delta }{\epsilon } \int _{0}^{t} |y_{1}(r) - y_{2}(r)|dr \end{aligned} \end{aligned}$$

Taking expectation:

$$\begin{aligned} \begin{aligned}&E(F(t)) \le E(F(0)) + (g-2\theta m) \int _{0}^{t} E |x_{1}(r) - x_{2}(r)|dr \\&\quad - \frac{d_1\eta }{\epsilon } \int _{0}^{t} E |s_{1}(r) - s_{2}(r)|dr - \frac{\delta }{\epsilon } \int _{0}^{t} E |y_{1}(r) - y_{2}(r)|dr \\&\quad \implies E(F(t)) + (2\theta m -g) \int _{0}^{t} E |x_{1}(r) - x_{2}(r)|dr \\&\quad + \frac{d_1\eta }{\epsilon } \int _{0}^{t} E |s_{1}(r) - s_{2}(r)|dr \\&\quad + \frac{\delta }{\epsilon } \int _{0}^{t} E |y_{1}(r) - y_{2}(r)|dr \le F(0) < \infty \end{aligned} \end{aligned}$$

So, it is obvious that $\int _{0}^{t} E |x_{1}(r) - x_{2}(r)|dr < \infty$, $\int _{0}^{t} E |s_{1}(r) - s_{2}(r)|dr < \infty$ and $\int _{0}^{t} E |y_{1}(r) - y_{2}(r)|dr < \infty ,$

i.e., $E |x_{1}(t) - x_{2}(t)|, E |s_{1}(t) - s_{2}(t)|, E |y_{1}(t) - y_{2}(t)| \in L^{1}[0,\infty ).$

So, $\lim _{t\rightarrow \infty }E |x_{1}(t) - x_{2}(t)|= \lim _{t\rightarrow \infty }E |s_{1}(t) - s_{2}(t)|= \lim _{t\rightarrow \infty }E |y_{1}(t) - y_{2}(t)|=0.$

Hence, $\lim _{t\rightarrow \infty } \left\{ E |x_{1}(t) - x_{2}(t)| + E |s_{1}(t) - s_{2}(t)| \right. \left. + E |y_{1}(t) - y_{2}(t)|\right\} =0.$

Hence the theorem. $\square$