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Prey–predator–scavenger model with Monod–Haldane type functional response

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Abstract

In this paper, we have developed a prey, predator and scavenger interaction dynamical model. The positivity, boundedness and stability conditions of our proposed system have been derived. Hopf bifurcation analysis has been done theoretically with respect to half saturation constant in the absence of direct measure of inhibitory effects \((\phi _1)\). Also, we have numerically studied Hopf bifurcation with respect to direct effects of inhibitory effects \((\phi _2)\) related to Monod–Haldane type functional response, carrying capacity of prey \((\gamma )\), death rate of scavenger \((\mu )\), conservation rate of prey \((\alpha )\) and harvesting rate of predator \((\eta )\). It is observed that the intra-species competition can lead our proposed system towards stability. It is also found that the harvesting of predator can control the chaotic dynamics of our proposed system. The increase of death rate of scavenger species can be stabilized our proposed system. Finally, some numerical simulation results have been presented for better understanding the dynamics of our proposed model.

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Appendix

Appendix

Proof of Theorem 1

From the first equation of system (2), we have

$$\begin{aligned} \frac{dx}{x}=\phi (x,y,z)dt \end{aligned}$$

where \(\phi (x,y,z)=(1-\gamma x)-\frac{y}{a_1+\phi _1x+\phi _2x^2}-z\). Then integrating the above equation within the limit [0, t], we have

$$\begin{aligned} x(t)=x(0)e^{\int _0^t \phi (x,y,z) dt}>0, \forall t \end{aligned}$$

Again, from the second equation of system (2), we have

$$\begin{aligned} \frac{dy}{y}=\psi (x,y,z)dt \end{aligned}$$

where \(\psi (x,y,z)=\frac{x}{a_1+\phi _1x+\phi _2x^2}-\delta -\eta y\). Integrating above equation in [0, t], it is obtained that

$$\begin{aligned} y(t)=y(0)e^{\int _0^t \psi (x,y,z) dt}>0, \forall t \end{aligned}$$

Also, from the third equation of system (2), we have

$$\begin{aligned} \frac{dz}{z}=\chi (x,y,z)dt \end{aligned}$$

where \(\chi (x,y,z)=x+\beta y-\mu -\nu z\). Then, integrating above equation in [0, t], we have

$$\begin{aligned} z(t)=z(0)e^{\int _0^t \chi (x,y,z) dt}>0, \forall t \end{aligned}$$

Hence all solutions of system (2) are non-negative. \(\square \)

Proof of Theorem 2

Let us construct a function

$$\begin{aligned} w=x+y+\frac{1}{\alpha }z \end{aligned}$$

Differentiating the above equation with respect to time, it is obtained that

$$\begin{aligned} \frac{d w}{d t}= & {} \frac{d x}{dt}+\frac{dy}{dt}+\frac{1}{\alpha }\frac{dz}{dt}\\= & {} x(1-\gamma x)-\frac{xy}{a_1+\phi _1x+\phi _2x^2}-xz+\frac{xy}{a_1 +\phi _1x+\phi _2x^2}-\delta y-\eta y^2\\&+\,xz+\frac{\beta }{\alpha }yz-\frac{\mu }{\alpha }z-\frac{\nu }{\alpha }z^2\\= & {} x(1-\gamma x)-\delta y-\eta y^2+\frac{\beta }{\alpha }yz-\frac{\mu }{\alpha }z-\frac{\nu }{\alpha }z^2 \end{aligned}$$

Now, we introduce a positive constant \(\sigma >0\), then we have

$$\begin{aligned} \frac{d w}{d t}+\sigma w= & {} (1+\sigma )x-(\delta -\sigma )y-\frac{1}{\alpha }(\mu -\sigma )z-\gamma x^2-\left( \eta y^2-\frac{\beta }{\alpha }yz+\frac{\nu }{\alpha }z^2\right) \nonumber \\\le & {} (1+\sigma )x-\gamma x^2- \left( \eta y^2-\frac{\beta }{\alpha }yz+\frac{\nu }{\alpha }z^2\right) , \text{ if } \sigma =\min \{\delta ,\mu \} \end{aligned}$$
(4)

The term \((\eta y^2-\frac{\beta }{\alpha }yz+\frac{\nu }{\alpha }z^2)\) is positive definite if \(\beta ^2>4\alpha \eta \nu \). Then equation (4) reduces to the form

$$\begin{aligned} \frac{d w}{d t}+\sigma w\le & {} (1+\sigma )x-\gamma x^2 \end{aligned}$$
(5)

Let \(f(x)=(1+\sigma )x-\gamma x^2\), then the maximum value of the function f(x) at \(x=\frac{1+\sigma }{2\gamma }\) is \(\frac{(1+\sigma )^2}{4\gamma }\). Then we can write

$$\begin{aligned} \frac{d w}{d t}+\sigma w\le & {} \frac{(1+\sigma )^2}{4\gamma } \end{aligned}$$
(6)

Solving Eq. (6), using differential inequality [33], it is obtained that

$$\begin{aligned} 0<w\le \frac{(1+\sigma )^2}{4\sigma \gamma }(1-e^{-\sigma t})+w(0)e^{-\sigma t} \end{aligned}$$

As \(t\rightarrow \infty \), we have

$$\begin{aligned} 0<w\le \frac{(1+\sigma )^2}{4\sigma \gamma } \end{aligned}$$

Hence all solutions of system (2) are bounded. \(\square \)

Proof of Theorem 8

Let us choose a suitable Lyapunov function

$$\begin{aligned} V=(x-x^*)-x^*ln\frac{x}{x^*}+(y-y^*)-y^*ln\frac{y}{y^*} +\frac{1}{\alpha }\bigg \{(z-z^*)-z^*ln\frac{z}{z^*}\bigg \} \end{aligned}$$

Differentiating above equation with respect to time, it is obtained that

$$\begin{aligned} \frac{d V}{dt}=\bigg (\frac{x-x^*}{x}\bigg )\frac{dx}{dt} +\bigg (\frac{y-y^*}{y}\bigg )\frac{dy}{dt}+\frac{1}{\alpha } \bigg (\frac{z-z^*}{z}\bigg )\frac{dz}{dt} \end{aligned}$$

Using the set of differential equation (2) and definition of equilibrium points, we have

$$\begin{aligned} \frac{d V}{dt}= & {} (x-x^*)\bigg \{1-\gamma x-\frac{y}{a_1+\phi _1x+\phi _2x^2}-z-1+\gamma x^*+\frac{y^*}{a_1+\phi _1x^*+\phi _2{x^*}^2}+z^*\bigg \}\\&+\,(y-y^*)\bigg \{\frac{x}{a_1+\phi _1x+\phi _2x^2}-\delta -\eta y-\frac{x^*}{a_1+\phi _1x^*+\phi _2{x^*}^2}+\delta +\eta y^*\bigg \}\\&+\,\frac{1}{\alpha }(z-z^*)\bigg \{\alpha x+\beta y-\mu -\nu z-\alpha x^*-\beta y^*+\mu +\nu z^*\bigg \}\\= & {} (x-x^*)\bigg \{-\gamma (x-x^*)-(z-z^*)-\bigg (\frac{y}{a_1 +\phi _1x+\phi _2x^2}-\frac{y^*}{a_1+\phi _1x^*+\phi _2{x^*}^2}\bigg )\bigg \}\\&+\,(y-y^*)\bigg \{\bigg (\frac{x}{a_1+\phi _1x+\phi _2x^2} -\frac{x^*}{a_1+\phi _1x^*+\phi _2{x^*}^2}\bigg )-\eta (y-y^*)\bigg \}\\&+\,\frac{1}{\alpha }(z-z^*)\bigg \{\alpha (x-x^*)+\beta (y-y^*)-\nu (z-z^*)\bigg \} \end{aligned}$$

After simplification of the above equation, it is obtained that

$$\begin{aligned} \frac{d V}{dt}= & {} -\gamma (x-x^*)^2-\eta (y-y^*)^2-\frac{\nu }{\alpha }(z-z^*)^2 +\frac{\beta }{\alpha }(y-y^*)(z-z^*)\\&-\,(x^*y-xy^*)\bigg \{\frac{(x-x^*)[\phi _1+\phi _2(x+x^*)]}{(a_1+\phi _1x+\phi _2x^2)(a_1+\phi _1x^*+\phi _2{x*}^2)}\bigg \}\\\le & {} -\gamma (x-x^*)^2-\eta (y-y^*)^2-\frac{\nu }{\alpha } (z-z^*)^2+\frac{\beta }{\alpha }(y-y^*)(z-z^*) \end{aligned}$$

if \(\frac{y}{y^*}>\frac{x}{x^*}>1\) and \(\frac{z}{z^*}>1\) holds.

Then, we can write

$$\begin{aligned} \frac{d V}{dt}\le & {} - \left[ \gamma (x-x^*)^2+\eta (y-y^*)^2+\frac{\nu }{\alpha }(z-z^*)^2-\frac{\beta }{\alpha }(y-y^*)(z-z^*)\right] \\ \frac{d V}{dt}\le & {} 0, \text{ if } \beta ^2<4\alpha \eta \nu \end{aligned}$$

Hence the proof. \(\square \)

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Panja, P. Prey–predator–scavenger model with Monod–Haldane type functional response. Rend. Circ. Mat. Palermo, II. Ser 69, 1205–1219 (2020). https://doi.org/10.1007/s12215-019-00462-9

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